What are the solutions to the equation solvefor x,x^2+4x-8=0 ?

The solutions to the equation solvefor x,x^2+4x-8=0 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

Find the zeros of solvefor x,x^2+4x-8=0

The zeros of solvefor x,x^2+4x-8=0 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

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x, x^{2} + 4 x - 8 = 0

x = 2 (3 - 1), x = - 2 (1 + 3)

Solution steps

x^{2} + 4 x - 8 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 8) = 43

x_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 3}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 3}{2 \cdot 1}

x = \frac{- 4 + 4 3}{2 \cdot 1} : 2 (3 - 1)

x = \frac{- 4 - 4 3}{2 \cdot 1} : - 2 (1 + 3)

The solutions to the quadratic equation are:

x = 2 (3 - 1), x = - 2 (1 + 3)

X^{2} + 3 X + 1 = 0

4 y^{2} + 20 y - 11 = 0

- 6 - 2 r^{2} = - 26

2 x^{2} = 4 x - 6

1 - x^{2} = 0

What are the solutions to the equation solvefor x,x^2+4x-8=0 ?
The solutions to the equation solvefor x,x^2+4x-8=0 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))
Find the zeros of solvefor x,x^2+4x-8=0
The zeros of solvefor x,x^2+4x-8=0 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))