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

Refine

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + i, x = 1 - i

3 x^{2} - 10 x - 5 = 0

x^{2} + 7^{2} = 1 0^{2}

x, x^{2} + 20 x + y^{2} - 20 y + 200 = 0

x, y = 2 - x^{2}

7 x^{2} = 15 x

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