What are the solutions to the equation x^2+12x=-32 ?

The solutions to the equation x^2+12x=-32 are x=-4,x=-8

Find the zeros of x^2+12x=-32

The zeros of x^2+12x=-32 are x=-4,x=-8

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x^{2} + 12 x = - 32

x = - 4, x = - 8

Solution steps

x^{2} + 12 x = - 32

Move

32

to the left side

x^{2} + 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

x_{1} = \frac{- 12 + 4}{2 \cdot 1}, x_{2} = \frac{- 12 - 4}{2 \cdot 1}

x = \frac{- 12 + 4}{2 \cdot 1} : - 4

x = \frac{- 12 - 4}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 4, x = - 8

25 p^{2} - 40 p + 16

- 6 < x - 4 \leq 1

x^{2} + 3 x + 2 + x^{2} - 1 = 0

4 (x - 5) = 5

\frac{x}{3} = - 2

What are the solutions to the equation x^2+12x=-32 ?
The solutions to the equation x^2+12x=-32 are x=-4,x=-8
Find the zeros of x^2+12x=-32
The zeros of x^2+12x=-32 are x=-4,x=-8