What are the solutions to the equation 1x^2+2x-8=0 ?

The solutions to the equation 1x^2+2x-8=0 are x=2,x=-4

Find the zeros of 1x^2+2x-8=0

The zeros of 1x^2+2x-8=0 are x=2,x=-4

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

x = 2, x = - 4

Solution steps

1 \cdot x^{2} + 2 x - 8 = 0

Refine

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

y, (4 - x) y^{2} = x^{3}

3 x^{2} + 18 x + 20

x^{2} + 10 + 9 = 0

3 x^{2} - 7 x - 11 = 0

x^{2} + 14 x + 49 = 44

What are the solutions to the equation 1x^2+2x-8=0 ?
The solutions to the equation 1x^2+2x-8=0 are x=2,x=-4
Find the zeros of 1x^2+2x-8=0
The zeros of 1x^2+2x-8=0 are x=2,x=-4