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

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

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

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

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

s = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

s = 1

The solution to the quadratic equation is:

s = 1

2 x^{2} + 15 x + 7 = 0

3 z + 5 = 14

\frac{2}{( 1 + 2 ) ^{3}}

5 x + 4 = - 6

3 x - 2 \leq 2 x + 4

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