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

The solutions to the equation x^2+12=-8x are x=-2,x=-6

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

The zeros of x^2+12=-8x are x=-2,x=-6

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x^{2} + 12 = - 8 x

x = - 2, x = - 6

Solution steps

x^{2} + 12 = - 8 x

Move

8 x

to the left side

x^{2} + 12 + 8 x = 0

Write in the standard form

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

x^{2} + 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 12 = 4

x_{1, 2} = \frac{- 8 \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4}{2 \cdot 1}, x_{2} = \frac{- 8 - 4}{2 \cdot 1}

x = \frac{- 8 + 4}{2 \cdot 1} : - 2

x = \frac{- 8 - 4}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = - 2, x = - 6

- 14 x^{2} + 3 x - 2 = 0

x^{2} - 6 x + 3 = 0

4 w^{2} = - 10 w

x, f = (x - h)^{2} + k

- 4 x^{2} + 5 x - 9 = 0

What are the solutions to the equation x^2+12=-8x ?
The solutions to the equation x^2+12=-8x are x=-2,x=-6
Find the zeros of x^2+12=-8x
The zeros of x^2+12=-8x are x=-2,x=-6