Someone pls help I don’t really understand graphing

Answer:

2×3.14×r=75

6.28 r=75

r=75/6.28=11.943=12 inch (nearly)

Answer:

-4

+3

12

-0.5, 12.25

x = -0.5

Step-by-step explanation:

The x intercepts are the values of x when y = 0 ie the roots of the equation
[tex]-x^2 -x +12 = 0[/tex]

or

[tex]x^2 + x - 12 = 0[/tex]

We can re-write the above as:
(x+4) (x-3) = 0
This gives the two roots as x = -4 and x = +3. Leftmost (smallest) root is -4 and rightmost(largest) root is +3

y intercept is when x = 0. Plugging into the original equation, y value at x = 0 is 12

Vertex x value is given by the formula -b/2a where a is the coefficient of x^2 and b the coefficient of x

Here a = -1, b= -1 so vertex x value = - (-1)/(-1).2 = - 1/2 = -0.5

Plugging this value of x into the original function gives the vertex y value
[tex]y = - (0.5^{2}) - (-0.5) + 12\\= -0.25 + 0.5 + 12 = 12.25[/tex]

The line of symmetry is the vertical line corresponding to the vertex x value so line of symmetry is at x = -0.5

The graph of the quadratic function shows these values

Answer = #2 - 1/3^3

Step-by-step explanation:

The equation of the function is f(x) = x(x + 2)(x -1)(x -3)

From the graph, we have the following highlight:

The above means that:

f(x) approaches positive infinity irrespective of the x value

Hence, the end behavior of the function is:

[tex]\mathrm{as}\:x\to \:+\infty \:,\:f\left(x\right)\to \:+\infty \:,\:\:\mathrm{and\:as}\:x\to \:-\infty \:,\:f\left(x\right)\to \:+\infty \:[/tex]

Since, the function opens upward.

Then the sign of the leading coefficient is positive

This is the point, where the graph crosses the x-axis.

From the graph, we have the zeros to be:

x = -2, x = 0, x = 1 and x = 3

Since the graph crosses the x-axis at this point, then the multiplicity of the zeros is 1

In (c), we have:

x = -2, x = 0, x = 1 and x = 3

Rewrite as:

x + 2= 0, x = 0, x - 1 = 0 and x - 3 = 0

Multiply these values

f(x) = x(x + 2)(x -1)(x -3)

Hence, the equation of the function is f(x) = x(x + 2)(x -1)(x -3)

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Your question is incomplete. Below you will find the missing content.

Which series of transformations will not map figure H onto itself?

A. O(x+0, y-2), reflection over y = 1

B. O(x+2, y-0), reflection over x=3

C. O(x+3, y+3), reflection over y = -x +7

D. O(x-3, y-3), reflection over y = -x +2

The correct option is Option D: the series of transformation O(x-3, y-3), reflection over y = -x +2 will not map figure H onto itself.

Here given a square with its 4 vertices at coordinates (2,1), (1,2), (2,3) and (3,2).

By checking all the options

Option A:

1st transformation (x+0, y-2) will map 4 vertices of the square into points of coordinates such as

(2,1) → (2,-1)

(1,2) → (1,0)

(2,3) → (2,1)

(3,2) → (3,0)

next transformation which is the reflection over y=1 gives the coordinates (x,2-y)

(2,-1) → (2,3)

(1,0) → (1,2)

(2,1) → (2,1)

(3,0) → (3,2)

These new points are exactly the same as the vertices of the initial square.

Option B:

1st transformation (x+2, y-0) will map 4 vertices of the square into points of coordinates such as

(2,1) → (4,1)

(1,2) → (3,2)

(2,3) → (4,3)

(3,2) → (5,2)

next transformation which is the reflection over x=3 gives the coordinates (6-x,y)

(4,1) → (2,1)

(3,2) → (3,2)

(4,3) → (2,3)

(5,2) → (1,2)

These new points are exactly the same as the vertices of the initial square.

Option C:

1st transformation (x+3, y+3) will map 4 vertices of the square into points of coordinates such as

(2,1) → (5,4)

(1,2) → (4,5)

(2,3) → (5,6)

(3,2) → (6,5)

next transformation which is the reflection over y=-x+7 gives the coordinates such as

(5,4) → (3,2)

(4,5) → (2,3)

(5,6) → (1,2)

(6,5) → (2,1)

These new points are exactly the same as the vertices of the initial square.

Option D:

1st transformation (x-3, y-3) will map 4 vertices of the square into points of coordinates such as

(2,1) → (-1,-2)

(1,2) → (-2,-1)

(2,3) → (-1,0)

(3,2) → (0,-1)

next transformation which is the reflection over y=-x+2 gives the coordinates such as

(-1,-2) → (4,3)

(-2,-1) → (3,4)

(-1,0) → (2,3)

(0,-1) → (3,2)

These new points are not matching with the vertices of the initial square.

Therefore the correct option is Option D: O(x-3, y-3), reflection over y = -x +2 will not map figure H onto itself.

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If the first inequality is written as:

3x ≤ -12

Then the correct option is D.

If it is written as it is in the problem, then neither of the graphs is the correct option.

Here we have the inequalities:

3x ≤ 12

2x + 3 > 11

(notice that in all the options, the first inequality has a negative sign, while in the two given inequalities there are no negative sign. This means that you will never get the correct answer if you use the given information, as the problem is incorrectly written).

First, we need to isolate x on both inequalities, we will get:

3x ≤ 12

x ≤ 12/3

x ≤ 4

(As you can see, because we don't have the negative sign, this inequality is different to all the ones in the options).

And for the other:

2x + 3 > 11

2x > 11 - 3

x > 8/2

x > 4

So neither of the options is actually correct.

If we instead write the first inequality as:

3x ≤ -12

We would solve this as:

x ≤ -4

And in that case, our two inequalities are:

x ≤ -4

x > 4

So the correct graph is graph D.

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Answer:

m = -3

c = 2

Step-by-step explanation:

This equation is given in slope-intercept form. The general structure of these equations is:

y = mx + c

In this form, "m" represents the slope and "c" represents the y-intercept. As you have been given the entire equation, the only thing you have to do is rearrange the equation to find which values correspond with the variables.

y = 2 - 3x  -----> y = -3x + 2

Therefore,

m = -3

c = 2

Answer:

y = -0.5x - 3.5

Explanation:

Take two points: (1, -4), (3, -5)

Find slope:

[tex]\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points[/tex]

[tex]\rightarrow \sf slope \ (m) : \dfrac{-5-(-4)}{3-1} = -\dfrac{1}{2}[/tex]

Equation:

[tex]\sf y- y_1 = m (x - x_1)[/tex]

[tex]\rightarrow \sf y- (-4) = -\dfrac{1}{2} (x -1)[/tex]

[tex]\rightarrow \sf y+ 4 = -\dfrac{1}{2} x + \dfrac{1}{2}[/tex]

[tex]\rightarrow \sf y = -\dfrac{1}{2} x + \dfrac{1}{2} - 4[/tex]

[tex]\rightarrow \sf y = -0.5 x -3.5[/tex]

The formula of a line is:

[tex]f(x) = kx + b[/tex]

k — stands for tg of line angle
b — stands for the y value of a point where the line crosses y axis.

We can see 2 points:
A (1, -4)
B (3, -5)

We can make a system of equations:

[tex]k*1+b=-4\\k*3+b=-5\\[/tex]

From second equation:

[tex]b = -5 - 3k[/tex]

Let's put b in first equation:

[tex]k-5-3k=-4\\-2k=1\\k=- \frac{1}{2}[/tex]

Now we can find b:

[tex]b = -5 -3*- \frac{1}{2}\\b = -5 + 1.5\\b = -3.5[/tex]

The answer is:
-0.5x -3.5

Answer:

c ×=4, only

Step-by-step explanation:

4*-8×+16=0

16-32=-16

-16=-16

Answer:

D. x=4 only

Step-by-step explanation:

You can solve using the quadratic equation which is [tex]x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex]

where in this case a=1, b=-8, and c=16. But there is any easier way to solve the equation. You can solve by completing the square which consists of moving the constant to the right side and then adding [tex](\frac{b}{2})^2[/tex]. To both sides that way you can rewrite the left side as a perfect square binomial. This is because [tex](x+b)^2[/tex] will expand to [tex]x^2 + 2xb + b^2[/tex]. So when you add [tex](\frac{b}{2})^2[/tex] to both equations. You're able to rewrite the left side as [tex](x+(\frac{b}{2}))^2[/tex]. Because once you expand it out, it becomes [tex]x^2 + 2(\frac{b}{2})x + (\frac{b}{2})^2[/tex]. Which simplifies to [tex]x^2 + bx + (\frac{b}{2})^2[/tex] which is the original equation. In this case there is no need to move the constant, which in this case is 16, to the other side. Since if you calculate the value of (-8/2)^2 it's equal to 16! So this can already be written as a perfect square binomial.

So the equation becomes

[tex]0 = (x+\frac{b}{2})^2[/tex]

Plug values in

[tex]0 = (x+(-\frac{8}{2}))^2[/tex]

Simplify

[tex]0 = (x-4)^2[/tex]

Take the square root of both sides

[tex]0 = x-4[/tex]

Add 4 to both sides

[tex]4 = x[/tex]

The solution set is 1 ≥ f ≥ -4, and the correct number line is the one in option A.

Here we have the inequality:

-1 ≤ -2f + 1 ≤ 9

To solve this, we need to isolate the "f" on the middle.

First, we can subtract 1 in both sides to get:

-2  ≤ -2f ≤ 8

Now we need to divide both sides by -2, remember that this will change the direction of the symbols:

-2/-2 ≥ f ≥ 8/-2

1 ≥ f ≥ -4

So we need to have a line that starts at -4 and ends at 1, the correct option is number line A.

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First term f = 1

If first four terms are f, f + d, f + 2d, f + 3d

f + f + d + f + 2d + f + 3d = 100

4f + 6d = 100 (divided by 2 , both sides )

2f + 3d = 50

2 + 3d = 50

3d = 50 – 2

3d = 48

d = 48/3

d = 16

The arithmetic sequence is 1, 17, 33, 49, ………….

Answer:

[tex]\sf\large\blue{\underbrace{\red{itz \: jass*}}}:[/tex]

Step-by-step explanation:

[tex]\sf\large\green{\underbrace{\red{hlo \: \: sat \: shri \: akal \: ji \: \: \: *}}}:[/tex]

[tex]{\huge \underline{{ \fbox \color{red}{A}}{\fbox \color{green}{n}}{\fbox \color{purple}{s}}{\fbox \color{brown}{w}}{\fbox \color{yellow}{e}}{\fbox \color{gray}{r } }}}[/tex]

[tex] \sf \dfrac{2}{15} \div \dfrac{3}{20} = \dfrac{2 \times 20}{15 \times 3} = \dfrac{40}{45} = \dfrac{8}{9} [/tex]

Answer :-

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \Rrightarrow \dfrac{8}{9} [/tex]

I hope I've helped : )

[tex]\dfrac{1}{50}[/tex]

[tex] \cfrac{2}{15} \times \cfrac{3}{20} [/tex]

[tex] \cfrac{2 \times 3}{15 \times 20} [/tex]

[tex] \cfrac{6}{15 \times 20} [/tex]

[tex] \cfrac{6}{300} [/tex]

[tex] \cfrac{1}{50} [/tex]

-------------------------------------------------------------------------------------------------------

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Answer:

x < 3

Step-by-step explanation:

The last step is to divide by -2. When you divide by a negative number you must reverse (flip) the inequality symbol.

-6 < -2x

-6/-2 > -2x/-2

3 > x

This is the same as

x < 3 (see the pointy side pointing at the x and the wide open side facing the 3)

Answer:

4

Step-by-step explanation:

In triangle JKL, the side opposite the 30 degree angle, JL, is half the length of the hypotenuse, KL. Thus, JL has a length of

[tex]4 \sqrt{2} [/tex]

In triangle JML, the hypotenuse is

[tex] \sqrt{2} [/tex]

times the length of the leg, meaning that x = 4

Answer:

111 cm^2

Step-by-step explanation:

h = 11.4

d = 5  so  r (radius) = 2.5

Surface Area of a Cone Formula: πr^2 + πrs

s represents the slant height of the cone, lets represent this with a drawing.

We have the values of h and r, but we don't have the value of s, meaning we have to find it. We can do this by using pythagorean theorem.

a^2 + b^2 = c^2

These three values can come together to create a right triangle which we can seperate from the rest of the diagram, now we can find s.

s = √(11.4^2 + 2.5^2)   so  s = 11.67

now that we have all of the pieces, we can plug back into the original surface area formula. πr^2 + πrs

π(2.5^2) + π(2.5)(11.67) = 111.29 cm^2 or 111 cm^2

The surface area of the solid cone requiring paint will be 111 square cm.

Let d be the diameter of the base circle, l be the slant height, and h be the height of the cone.

Then the lateral surface area of the cone will be

SA = πrl + πr²

Then the slant height is given as

l² = r² + h²

The exterior of a solid cone is painted. The height of the cone is 11.4 centimeters, and the diameter of its opening is 5 centimeters.

Then the radius of the base circle will be

r = 5 / 2

r = 2.5 cm

The length of the slant height will be

l² = 11.4² + 2.5²

l² = 136.21

 l = 11.67

Then the surface area of the solid cone requiring paint will be

SA = π × 2.5 × 11.67 + π × 2.5²

SA = 29.18π + 6.25π

SA = 35.43π

SA = 111.3

SA ≈ 111 square cm

The surface area of the solid cone requiring paint will be 111 square cm.

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Answer:

65

Step-by-step explanation:

its simple math

Answer:

x+5=12.          x = 7,

7x=21.            x = 3,

b-2/3=5/6.     b = 1.5,

3/4x=6x=8.    x = 1/3,

45=3(x+1).      x = 14,

22.5+1.5x.      x = 15,

20+(5•9).        $65

Step-by-step explanation:

Answer:

  (c)  (-30, 10)

Step-by-step explanation:

Two points will be on the same graph of a proportional relationship if the ratio of values y/x is the same.

__

The given point (x, y) = (60, -20) has a ratio k = y/x = -20/60 = -1/3.

The offered points have y/x ratios ...

(a) 30/-10 = -3

(b) -15/30 = -1/2

(c) 10/-30 = -1/3 . . . . . . the point you are looking for is (-30, 10)

(d) -30/80 = -3/8

Answer:

opotion has not correct answer

answer is 15.

Answer:

[tex]3^\frac{7}{10}[/tex]    or      [tex]\sqrt[10]{3^7}[/tex]

Step-by-step explanation:

[tex]\left(\sqrt{3}\right)\left(\sqrt[5]{3}\right)=\sqrt[10]{3^7}\quad[/tex]

(√3)([tex]\sqrt[5]{3}[/tex]) = √3 · [tex]\sqrt[5]{3}[/tex]

{√3 = [tex]3^{\frac{1}{2}}[/tex]}                 {radical rule: [tex]\sqrt{x}=x^1^/^2[/tex]}

[tex]\sqrt3[/tex][tex]\sqrt[5]{3}[/tex] =  [tex]3^{\frac{1}{2}}[/tex] · [tex]\sqrt[5]{3}[/tex]

{[tex]\sqrt[5]{3}[/tex] = [tex]3^{\frac{1}{5}}[/tex]}                  {radical rule: [tex]\sqrt[n]{x} = x^1^/^n[/tex]}

 [tex]3^{\frac{1}{2}}[/tex] · [tex]\sqrt[5]{3}[/tex] =   [tex]3^{\frac{1}{2}}[/tex] · [tex]3^{\frac{1}{5}}[/tex]     {exponent rule:  [tex]a^x*a^y=a^x^+^y[/tex]}

                (1/2 + 1/5 = 5/10 + 2/10 = 7/10)

              [tex]=3^\frac{7}{10}[/tex]        {opposite of radical rule: [tex]\sqrt[n]{x} = x^1^/^n[/tex] ;  [tex]x^\frac{a}{b}=\sqrt[b]{x^a}[/tex]}

             = [tex]\sqrt[10]{3^7}[/tex]

so, the simplified version of this equation can either be written as:

[tex]3^\frac{7}{10}[/tex]    or      [tex]\sqrt[10]{3^7}[/tex]

hope this helps!!

(I can't clearly see the last option, but if it's either of these, then it's correct)

Answer:

[tex]\fbox {400.41 feet}[/tex]

Step-by-step explanation:

Here we need to take the tan ratio of the angle.

⇒ tan 4° = 28/x = 0.0699268119

⇒ x = 28/0.0699268119

⇒ x = 400.41 feet

Answer:

t+s = 9

4t+9s=51

tulips and succulents combined is 9 plants

the amount of money they have combined is $51

Answer:

134

Step-by-step explanation:

the "nth" term of a sequence means that you can plug in the term number to get the term value

so, if

  11n + 2 is the sequence

  11(12) + 2 is the 12th term

  132 + 2 is the 12th term

  134 is the 12th term

By plugging in the term "place" (1st, 2nd, 3rd, 4th, etc.) for the variable "n", we get a term value.

hope this helps!!

The 12th term of the sequence is 133

The nth term is given as:

Tn = 11n + 1

The 12th term is calculated as:

T(12) = 11(12) + 1

Evaluate

T(12) = 133

Hence, the 12th term of the sequence is 133

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Answer:

3/5 ..................................  6/9

Step-by-s4tep explanation:

Answer:

300 meters per minute

Step-by-step explanation:

Step 1

Determine the fraction of the total distance for each part of the race.

Given:

[tex]\begin{aligned} \implies \sf 2nd\:part & = \sf \dfrac{2}{5}\:of\:remaining\:distance\\ & = \sf \dfrac{2}{5} \times\left(1-\dfrac{4}{9}\right)\\& = \sf \dfrac{2}{9}\:of\:total\:distance\end{aligned}[/tex]

[tex]\begin{aligned}\implies \sf 3rd\:part & = \sf 1-1st\:part-2nd\:part\\& = \sf 1-\dfrac{4}{9}-\dfrac{2}{9}\\& = \sf \dfrac{1}{3}\:of\:total\:distance\end{aligned}[/tex]

Step 2

Determine the distance of the third part of the race.

Given:

[tex]\begin{aligned}\sf Distance & = \sf speed \times time\\& = \sf 300 \times 15\\& = \sf 4500 \:\:meters\end{aligned}[/tex]

Step 3

If the third part of the race (which is 1/3 of the total distance) is 4500m, then the distance of the whole race is:

[tex]\begin{aligned} \sf Total\:distance & = \sf 4500 \times 3\\& = \sf 13500\:meters \end{aligned}[/tex]

Step 4

Determine the distance of the 1st part of the race:

[tex]\begin{aligned}\sf Distance\:of\:1st\:part\:of\:race & = \sf \dfrac{4}{9} \:of\:total\:distance\\& = \sf \dfrac{4}{9} \times 13500\\& = \sf 6000\:meters \end{aligned}[/tex]

Step 5

Determine the speed of the 1st part of the race:

Given:

[tex]\begin{aligned}\sf Speed\:of\:1st\:part\:of\:race & = \sf \dfrac{distance}{time}\\& = \sf \dfrac{6000}{20}\\ & = \sf 300\:meters\:per\:minute\end{aligned}[/tex]

Answer:

Slope is 3/4

Step-by-step explanation:

Gradient means slope. So, to find the slope without using the formula, look at two points then count up until you're in line with the other point, then count right until you reach the other point. Remember it must be written as rise/run.

I hope this explanation makes sense!

The points I will use is (0,-2) and (4,1).

Slope is 3/4.

Hope this helps!

If not, I am sorry.

Answer:

cbrt(x+1)-3=y

Step-by-step explanation:

The inverse of a function is when the domain and range are swapped. So to find the inverse simply swap the x and y in the equation and then solve for y.

[tex]y = (x+3)^3-1\\x=(y+3)^3-1\\x+1=(y+3)^3\\\sqrt[3]{x+1}=y+3\\\sqrt[3]{x+1}-3=y[/tex]

Answer:

C.

Step-by-step explanation:

The reds are counting up in whole numbers by one integer.

Hope this helps!

If not, I am sorry.

Use the chain rule to differentiate both sides of the given equation with respect to [tex]x[/tex] :

[tex]y = 2 f(g(x))[/tex]

[tex]\dfrac{dy}{dx} = y' = 2 f'(g(x)) g'(x)[/tex]

Differentiate again with the product and chain rules to get the second derivative :

[tex]\dfrac{d^2y}{dx^2} = y'' = \boxed{2 f'(g(x)) g''(x) + 2 f''(g(x)) g'(x)^2}[/tex]

(D)

Answer:

[tex]x=\dfrac{11}{4}[/tex]

Step-by-step explanation:

Given:

[tex]9^{-\frac{1}{2}}=\dfrac{27^{\frac{1}{4}}}{3^{(x-1)}}[/tex]

Rewrite 9 as 3² and 27 as 3³:

[tex]\implies (3^2)^{-\frac{1}{2}}=\dfrac{(3^3)^{\frac{1}{4}}}{3^{(x-1)}}[/tex]

[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]

[tex]\implies 3^{-1}=\dfrac{3^{\frac{3}{4}}}{3^{(x-1)}}[/tex]

Multiply both sides by [tex]3^{(x-1)}[/tex] :

[tex]\implies 3^{-1} \cdot 3^{(x-1)}=3^{\frac{3}{4}}[/tex]

[tex]\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c}:[/tex]

[tex]\implies 3^{(-1+x-1)}=3^{\frac{3}{4}}[/tex]

[tex]\implies 3^{(x-2)}=3^{\frac{3}{4}}[/tex]

[tex]\textsf{Apply exponent rule} \quad a^{f(x)}=a^{g(x)} \implies f(x)=g(x):[/tex]

[tex]\implies x-2=\dfrac{3}{4}[/tex]

[tex]\implies 4(x-2)=3[/tex]

[tex]\implies 4x-8=3[/tex]

[tex]\implies 4x=11[/tex]

[tex]\implies x=\dfrac{11}{4}[/tex]

The value of the expression 2 2/3 - (-3 2/6) is 6

The expression is given as:

2 2/3 - (-3 2/6)

Open the bracket

2 2/3 + 3 2/6

Express as improper fractions

8/3 + 20/6

Rewrite as:

16/6 +20/6

Evaluate the sum

36/6

Evaluate the quotient

6

Hence, the value of 2 2/3 - (-3 2/6) is 6

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The t-test might be more useful in actual practice when standard deviation or variance is unknown.

The t-test is used to determine how the averages of various data sets differ from one another.

When variance is not provided, the T-test, a particular kind of parametric test, is used to determine how the means of two sets of data differ from one another. When variance is provided, the Z-test indicates a hypothesis test that determines whether the means of two datasets differ from one another.

Example of t-test: If you flip a coin 1,000 times, for instance, you can discover that the outcome is distributed normally over all trials.

Example of z-test: We are conducting research using information gathered from cohorts of students who have taken Elementary Statistics in the past.

T-tests are typically more suitable when addressing issues with small sample sizes, whereas z-tests are suitable for large sample sizes.

Using the t-test for an ordinal variable, their frequency is not even close to a normal distribution, and the arithmetic mean offers an unsuitable measure of location.

Because we frequently don't know the population standard deviation, Z-Tests have this drawback.

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Answer:

[tex]\large {\textsf{B and E}}\ \implies x_1=-2\sqrt{5}-3,\ x_2=+2\sqrt{5}-3[/tex]

Step-by-step explanation:

Quadratic Formula: [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

Quadratic Equation: ax² + bx + x = 0, where a ≠ 0

Given equation: x² + 6x + 9 = 20

Step 1: Subtract 20 from both sides.

x² + 6x + 9 - 20 = 20 - 20

⇒ x² + 6x - 11 = 0

Step 2: Identify the values of a, b, and c and substitute them in the formula.

⇒ a = 1, b = 6, c = -11

[tex]x=\dfrac{-6\pm\sqrt{\bold{6^2}-4(1)(-11)}}{\bold{2(1)}}\\\\x=\dfrac{-6\pm\sqrt{36\bold{\ - \ 4(-11)}}}{2}\\\\x=\dfrac{-6\pm\sqrt{\bold{36+44}}}{2}\\\\x=\dfrac{-6\pm\sqrt{80}}{2}\\\\x=\dfrac{-6\pm\sqrt{16\times5}}{2}\implies \dfrac{-6\pm\sqrt{\bold{16}}\times\sqrt{5}}{2} \\\\x=\dfrac{-6\pm4\sqrt{5}}{2}[/tex]

Step 3: Separate into possible cases.

[tex]x_1=\dfrac{-6-4\sqrt{5}}{2}\implies \dfrac{-6}{2}+\dfrac{-4\sqrt{5}}{2}\implies \boxed{-3-2\sqrt{5}\ \ \textsf{or}\ -2\sqrt{5}-3}\\\\x_2=\dfrac{-6+4\sqrt{5}}{2}\implies \dfrac{-6}{2}+\dfrac{4\sqrt{5}}{2}\implies \boxed{-3+2\sqrt{5}\ \ \textsf{or}\ \ 2\sqrt{5}-3}[/tex]

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