2(y+3)^2-3y^2-2(y+3)+y=30

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Solution for 2(y+3)^2-3y^2-2(y+3)+y=30 equation: 



2(y+3)^2-3y^2-2(y+3)+y=30

We move all terms to the left:

2(y+3)^2-3y^2-2(y+3)+y-(30)=0

We add all the numbers together, and all the variables

-3y^2+y+2(y+3)^2-2(y+3)-30=0

We multiply parentheses

-3y^2+y+2(y+3)^2-2y-6-30=0

We add all the numbers together, and all the variables

-3y^2-1y+2(y+3)^2-36=0

We move all terms containing y to the left, all other terms to the right

-3y^2-1y+2(y+3)^2=36

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