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

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

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

The zeros of x^2-2x+2=1 are x=1

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

x = 1

Solution steps

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

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

x, x^{2} y^{3} + 2 y = 6 x - 3

7 x^{2} - 11 x + 5 = 0

q, \frac{5 q ^{2}}{R} = \frac{B}{a}

2 x^{2} + 16 x + 28 = 0

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

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