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

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

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

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

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

x = 1 - i, x = 1 + i

Solution steps

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 (- 1) (- 2) : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - i, x = 1 + i

2 x^{2} + 7 x \geq 15

- \frac{1}{6} - \frac{1}{3}

∣ x - 2 ∣ - 3 = 6

\frac{g}{3} - 7 = 15

x^{2} y^{3}

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