Question

How do you solve -y +28 +y^2 =2y +2y^2 for y?

Answer 1

We are given the following:

Solve for y.

`-y+28+y^2=2y+2y^2`

This being said, lets begin.

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1. First, move
`2y^2` to the left side.

`-y+28-y^2=2y`

2. Move
`2y` to the left side.

`-y^2-3y+28=0`

3. Solve with the quadratic formula.

`y_(1,2) = (-(-3) -+ √((-3)^2-4(-1)*28) )/(2(-1))`

4.
`√((-3)^2-4(-1)*28)=11`

`y_(1,2) =(-(3)-+11)/(2(-1))`

5. Separate the solutions.

`y=(-(-3)+11)/(2(-1)) , y_(2) =(-(-3)-11)/(2(-1))`

6.

`y=(-(-3)+11)/(2(-1))`
`=-7` ← First Solution

7.

`y=(-(-3)-11)/(2(-1)) =4` ←Second Solution

Final Solution:

`y=-7,y=4`

Hope this helps!