4(y2-3y)-8=2y(2y+4)+32

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Solution for 4(y2-3y)-8=2y(2y+4)+32 equation: 



4(y2-3y)-8=2y(2y+4)+32

We move all terms to the left:

4(y2-3y)-8-(2y(2y+4)+32)=0

We add all the numbers together, and all the variables

4(+y^2-3y)-(2y(2y+4)+32)-8=0

We multiply parentheses

4y^2-12y-(2y(2y+4)+32)-8=0



We calculate terms in parentheses: -(2y(2y+4)+32), so:

2y(2y+4)+32

We multiply parentheses

4y^2+8y+32

Back to the equation:

-(4y^2+8y+32)



We get rid of parentheses

4y^2-4y^2-12y-8y-32-8=0

We add all the numbers together, and all the variables

-20y-40=0

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

-20y=40

y=40/-20

y=-2

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