What are the solutions to the equation factor-y+28+y^2=2y+2y^2 ?

The solutions to the equation factor-y+28+y^2=2y+2y^2 are y=4,y=-7

Find the zeros of factor-y+28+y^2=2y+2y^2

The zeros of factor-y+28+y^2=2y+2y^2 are y=4,y=-7

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

y = 4, y = - 7

Solution steps

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

Move

2 y^{2}

to the left side

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

Move

2 y

to the left side

- y^{2} - 3 y + 28 = 0

Factor

- y^{2} - 3 y + 28 : - (y - 4) (y + 7)

- (y - 4) (y + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 4 = 0 or y + 7 = 0

Solve

y - 4 = 0 : y = 4

Solve

y + 7 = 0 : y = - 7

The solutions to the quadratic equation are:

y = 4, y = - 7

16 x^{2} + 72 x + 83 = 0

k^{2} + 5 k - 3 = 0

x^{2} + 8 x + 15

3 (3 x + 5)^{2} + 10 = 22

5 x^{2} + 5 x - 9 = 0

What are the solutions to the equation factor-y+28+y^2=2y+2y^2 ?
The solutions to the equation factor-y+28+y^2=2y+2y^2 are y=4,y=-7
Find the zeros of factor-y+28+y^2=2y+2y^2
The zeros of factor-y+28+y^2=2y+2y^2 are y=4,y=-7