What are the solutions to the equation 8-x^2=4x+12 ?

The solutions to the equation 8-x^2=4x+12 are x=-2

Find the zeros of 8-x^2=4x+12

The zeros of 8-x^2=4x+12 are x=-2

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8 - x^{2} = 4 x + 12

x = - 2

Solution steps

8 - x^{2} = 4 x + 12

Move

12

to the left side

- x^{2} - 4 = 4 x

Move

4 x

to the left side

- x^{2} - 4 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - 4 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) (- 4) = 0

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 ( - 1 )}

x = \frac{- ( - 4 )}{2 ( - 1 )}

\frac{- ( - 4 )}{2 ( - 1 )} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

5 x^{2} + 7 x^{2} - 25 5^{2} = 0

2 x^{2} - x - 60 = 0

y, y^{2} = \frac{x ^{3}}{2 - x}

x, (x + 3)^{2} + (y + 1)^{2} = r^{2}

4 x^{2} - 12 x = - 9

What are the solutions to the equation 8-x^2=4x+12 ?
The solutions to the equation 8-x^2=4x+12 are x=-2
Find the zeros of 8-x^2=4x+12
The zeros of 8-x^2=4x+12 are x=-2