What are the solutions to the equation ((y-2)^2)/2 =6-y ?

The solutions to the equation ((y-2)^2)/2 =6-y are y=4,y=-2

Find the zeros of ((y-2)^2)/2 =6-y

The zeros of ((y-2)^2)/2 =6-y are y=4,y=-2

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\frac{( y - 2 ) ^{2}}{2} = 6 - y

y = 4, y = - 2

Solution steps

\frac{( y - 2 ) ^{2}}{2} = 6 - y

Multiply both sides by

2

(y - 2)^{2} = 12 - 2 y

Expand

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

y^{2} - 4 y + 4 = 12 - 2 y

Move

2 y

to the left side

y^{2} - 2 y + 4 = 12

Move

12

to the left side

y^{2} - 2 y - 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

y = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

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

The solutions to the quadratic equation are:

y = 4, y = - 2

- 16 t^{2} + 32 t + 128

2 w^{2} + 5 w - 63 = 0

x, f (a + h) = 7 x^{2} + 6

5 x^{4} + 10 x^{3} + 15 x^{2}

- 4 - 2 i

What are the solutions to the equation ((y-2)^2)/2 =6-y ?
The solutions to the equation ((y-2)^2)/2 =6-y are y=4,y=-2
Find the zeros of ((y-2)^2)/2 =6-y
The zeros of ((y-2)^2)/2 =6-y are y=4,y=-2