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

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

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

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

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

y = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

y = 1

The solution to the quadratic equation is:

y = 1

3 x - x + 4 = - 4

x^{3} + 3 x^{2} + 6 x + 8

c^{2} - 9

\frac{a}{4} = 3 - \frac{a - 3}{5}

\frac{4}{3} - 3

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