What are the solutions to the equation 6-y=((y-3)^2)/4 ?

The solutions to the equation 6-y=((y-3)^2)/4 are y=5,y=-3

Find the zeros of 6-y=((y-3)^2)/4

The zeros of 6-y=((y-3)^2)/4 are y=5,y=-3

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6 - y = \frac{( y - 3 ) ^{2}}{4}

y = 5, y = - 3

Solution steps

6 - y = \frac{( y - 3 ) ^{2}}{4}

Multiply both sides by

4

24 - 4 y = (y - 3)^{2}

Expand

(y - 3)^{2} : y^{2} - 6 y + 9

24 - 4 y = y^{2} - 6 y + 9

Switch sides

y^{2} - 6 y + 9 = 24 - 4 y

Move

4 y

to the left side

y^{2} - 2 y + 9 = 24

Move

24

to the left side

y^{2} - 2 y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 15) = 8

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

Separate the solutions

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

y = \frac{- ( - 2 ) + 8}{2 \cdot 1} : 5

y = \frac{- ( - 2 ) - 8}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

y = 5, y = - 3

e^{2} + e^{- 2}

200 x^{2} - 50

5 x - 1 = 12 x - 57

lo g_{4} (4)

(x y)^{- 1} (x + y)

What are the solutions to the equation 6-y=((y-3)^2)/4 ?
The solutions to the equation 6-y=((y-3)^2)/4 are y=5,y=-3
Find the zeros of 6-y=((y-3)^2)/4
The zeros of 6-y=((y-3)^2)/4 are y=5,y=-3