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

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

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

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

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

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

Decimal Notation

x = 1.17157 \dots, x = 6.82842 \dots

Solution steps

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

Solve with the quadratic formula

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

8^{2} - 4 (- 1) (- 8) = 42

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

Separate the solutions

x_{1} = \frac{- 8 + 4 2}{2 ( - 1 )}, x_{2} = \frac{- 8 - 4 2}{2 ( - 1 )}

x = \frac{- 8 + 4 2}{2 ( - 1 )} : 2 (2 - 2)

x = \frac{- 8 - 4 2}{2 ( - 1 )} : 2 (2 + 2)

The solutions to the quadratic equation are:

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

p^{2} + 10 p + 1 = 0

x^{2} = \frac{5}{2}

2 (3 x - 1)^{2} = 32

16 - x^{2} = 16 - 8 x + x^{2}

(x + 1)^{2} - 9 (x + 1) + 20 = 0

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