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

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

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

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

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

x = 1

Solution steps

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

2 - y = 2

y^{2} + 3 y - 3 = 0

15 - 0.06 x \geq 5

(x - 2) = x^{2} - 4

0 = 4 x^{2} + 2 x - 1

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