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

The solutions to the equation x^2+2x-8=-8 are x=0,x=-2

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

The zeros of x^2+2x-8=-8 are x=0,x=-2

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

x = 0, x = - 2

Solution steps

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

Move

8

to the left side

x^{2} + 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

x = \frac{- 2 + 2}{2 \cdot 1} : 0

x = \frac{- 2 - 2}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

- \frac{7}{8} > - 0.5

6 m^{3} - 12 mn + m^{2} n - 2 n^{2}

- 4 x^{2} + 7 x - 3 = 0

\frac{2}{\frac{2}{3}}

3 v^{2} + 10 v + 3

What are the solutions to the equation x^2+2x-8=-8 ?
The solutions to the equation x^2+2x-8=-8 are x=0,x=-2
Find the zeros of x^2+2x-8=-8
The zeros of x^2+2x-8=-8 are x=0,x=-2