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

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

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

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

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

x = - 2

Solution steps

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

Move

8

to the left side

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 2 \cdot 8 = 0

x_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 2}

x = \frac{- 8}{2 \cdot 2}

\frac{- 8}{2 \cdot 2} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

(2 k^{3} - 7 k^{2} + 3 k) - (- 4 k^{2} + 5 k)

p^{2} + 12 p + 36

- 3 + \frac{1}{3} x = - 8

9 c - 1 < 4 c - 5

t^{3} - 5 t^{2} + 10 - 2 t

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