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

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

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

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

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

x = 1 + i, x = 1 - i

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

2 x^{2} + 8 x + 10 = 0

x^{2} + 4 = 5 x + 4

5 x^{2} - 61 = - 16

(x + 5) (x - 5) = 11

2 x^{2} + 9 x - 5 = 0

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