What are the solutions to the equation (((y-4)^2))/4 =7-y ?

The solutions to the equation (((y-4)^2))/4 =7-y are y=6,y=-2

Find the zeros of (((y-4)^2))/4 =7-y

The zeros of (((y-4)^2))/4 =7-y are y=6,y=-2

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\frac{( ( y - 4 ) ^{2} )}{4} = 7 - y

y = 6, y = - 2

Solution steps

\frac{( ( y - 4 ) ^{2} )}{4} = 7 - y

Multiply both sides by

4

(y - 4)^{2} = 28 - 4 y

Expand

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

y^{2} - 8 y + 16 = 28 - 4 y

Move

4 y

to the left side

y^{2} - 4 y + 16 = 28

Move

28

to the left side

y^{2} - 4 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

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

The solutions to the quadratic equation are:

y = 6, y = - 2

2. 5^{2}

4 x^{2} - 1 = - 8 x

4 w^{3} + 7 w^{2} + 12 w + 21

2 \cdot 2^{n}

\frac{8}{12} (2 x - 1) + \frac{3}{4} (x + 1) = 3

What are the solutions to the equation (((y-4)^2))/4 =7-y ?
The solutions to the equation (((y-4)^2))/4 =7-y are y=6,y=-2
Find the zeros of (((y-4)^2))/4 =7-y
The zeros of (((y-4)^2))/4 =7-y are y=6,y=-2