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

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

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

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

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

x = - 4, x = 2

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 6}{2 ( - 1 )} : - 4

x = \frac{- ( - 2 ) - 6}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

x = - 4, x = 2

x^{2} + 20 x + 75 = 0

4 + 3 u = - 11

29 x - 25 < 91

2 y^{2} - 3 y - 2

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