What are the solutions to the equation (y-4)^2=8(y-2) ?

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

Find the zeros of (y-4)^2=8(y-2)

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

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(y - 4)^{2} = 8 (y - 2)

y = 4 (2 + 2), y = 4 (2 - 2)

Decimal Notation

y = 13.65685 \dots, y = 2.34314 \dots

Solution steps

(y - 4)^{2} = 8 (y - 2)

Expand

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

Expand

8 (y - 2) : 8 y - 16

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

Move

16

to the left side

y^{2} - 8 y + 32 = 8 y

Move

8 y

to the left side

y^{2} - 16 y + 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 32 = 82

y_{1, 2} = \frac{- ( - 16 ) \pm 8 2}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 16 ) + 8 2}{2 \cdot 1}, y_{2} = \frac{- ( - 16 ) - 8 2}{2 \cdot 1}

y = \frac{- ( - 16 ) + 8 2}{2 \cdot 1} : 4 (2 + 2)

y = \frac{- ( - 16 ) - 8 2}{2 \cdot 1} : 4 (2 - 2)

The solutions to the quadratic equation are:

y = 4 (2 + 2), y = 4 (2 - 2)

m^{2} - 14 m - 239 = 0

\frac{x ^{2}}{4} + \frac{3 x}{2} + 2 = 0

4 x^{2} + 6 x + 7 = 0

x^{2} + 100 = 20 x

x^{2} - 7 x + 29 = 3 x + 4

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