# Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

x² – 21 = –4x
• 7, 3
• –7, 3
• 7, –3
• –7, –3

## Why did the quadratic equation go to the circus? It wanted to solve for the clown roots! Let's use the quadratic formula to find those clown roots for this equation.

The quadratic formula is x = (-b ± √(b² - 4ac))/(2a)

For this equation, let's rearrange it so it's in standard form ax² + bx + c = 0. We have x² + 4x - 21 = 0. In this case, a = 1, b = 4, and c = -21.

Plugging these values into the quadratic formula, we have:

x = (-4 ± √(4² - 4(1)(-21)))/(2(1))

x = (-4 ± √(16 + 84))/2

x = (-4 ± √100)/2

x = (-4 ± 10)/2

Now we can simplify this further:

x = (-4 + 10)/2 = 6/2 = 3

x = (-4 - 10)/2 = -14/2 = -7

So the clown roots for this equation are 3 and -7.

## To use the quadratic formula, we need to write the equation in the standard form: ax^2 + bx + c = 0.

Given equation: x² – 21 = –4x

Rearranging the terms:
x² + 4x - 21 = 0

Now, we can directly substitute the values of a, b, and c into the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

a = 1, b = 4, c = -21

Substituting the values:
x = (-4 ± √(4² - 4(1)(-21))) / (2(1))

Simplifying:
x = (-4 ± √(16 + 84)) / 2
x = (-4 ± √100) / 2
x = (-4 ± 10) / 2

Now, we solve for both possible values of x using the plus-minus symbol (±):

For x = (-4 + 10) / 2:
x = 6 / 2
x = 3

For x = (-4 - 10) / 2:
x = -14 / 2
x = -7

Therefore, the solutions to the equation x² - 21 = -4x are:
x = 3 and x = -7

• 7, 3
• -7, 3

## put it into standard form of a quadratic ...

x^2 + 4x - 21 = 0
how nice that it factors ....

(x+7)(x-3) = 0

so x = ... or x = ....

## But you wanted to use the formula ...

x = (-4 ± √100)/2
= (-4 ± 10)/2
= (-4+10)/2 or (-4-10)/2
= .....