What are the solutions to the equation (y-4)^2=2y ?

The solutions to the equation (y-4)^2=2y are y=8,y=2

Find the zeros of (y-4)^2=2y

The zeros of (y-4)^2=2y are y=8,y=2

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

y = 8, y = 2

Solution steps

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

Expand

(y - 4)^{2} : y^{2} - 8 y + 16

y^{2} - 8 y + 16 = 2 y

Move

2 y

to the left side

y^{2} - 10 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 16 = 6

y_{1, 2} = \frac{- ( - 10 ) \pm 6}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 6}{2 \cdot 1}, y_{2} = \frac{- ( - 10 ) - 6}{2 \cdot 1}

y = \frac{- ( - 10 ) + 6}{2 \cdot 1} : 8

y = \frac{- ( - 10 ) - 6}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

y = 8, y = 2

x^{2} + 6 x - 135 = 0

(2 - x)^{2} = 36

a, an = 1 + n - n^{2}

(2 x - 5)^{2} = (x - 6)^{2} + (2 x + 1) (2 x - 9)

x^{2} = \frac{1}{100}

What are the solutions to the equation (y-4)^2=2y ?
The solutions to the equation (y-4)^2=2y are y=8,y=2
Find the zeros of (y-4)^2=2y
The zeros of (y-4)^2=2y are y=8,y=2