What are the solutions to the equation 1x^2-2x+2=0 ?

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

Find the zeros of 1x^2-2x+2=0

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

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

x = 1 + i, x = 1 - i

Solution steps

1 \cdot 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

- x^{2} + 144 = 0

2 x^{2} - 3 x + 1

x^{2} + 8 x - 192 = 0

(x - 5) (x - 1) (x + 2) + x (2 x - 11) (x + 2) = x (3 x - 12) (x - 1)

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

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