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

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

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

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

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

x = 2 - 2, x = 2 + 2

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 2) = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 - 2, x = 2 + 2

(32 - 8)^{2}

1 \leq y < 4

- 2 x + 5 = 13

\frac{\frac{2}{9}}{( - \frac{6}{5} )}

56 (23)

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