What are the solutions to the equation y^2-4y+4=0 ?

The solutions to the equation y^2-4y+4=0 are y=2

Find the zeros of y^2-4y+4=0

The zeros of y^2-4y+4=0 are y=2

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y^{2} - 4 y + 4 = 0

y = 2

Solution steps

y^{2} - 4 y + 4 = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

y = \frac{- ( - 4 )}{2 \cdot 1}

\frac{- ( - 4 )}{2 \cdot 1} = 2

y = 2

The solution to the quadratic equation is:

y = 2

\frac{8 - 4 x}{10} = 4 + 2 x

25 t^{2} - 64

1 + 7 x - 3 > 3

6 + x = - 54

t^{2} + 7 t - 1 = 0

What are the solutions to the equation y^2-4y+4=0 ?
The solutions to the equation y^2-4y+4=0 are y=2
Find the zeros of y^2-4y+4=0
The zeros of y^2-4y+4=0 are y=2