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

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

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

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

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roots

x^{2} + 2 x - 8

x = 2, x = - 4

Solution steps

x^{2} + 2 x - 8

The roots are the intercepts with the x-axis

(y = 0)

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 6}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2, x = - 4

x^{2} + 2 x - 3

x^{2} - 14 x + 196 = 0

- 2 x^{2} - 4 x + 7

3 x^{2} + 12 x - 7 = 0

3 x^{2} - 4 x - 12 = 10

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