2y(y+3)=(y+4)(2y-1)

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Solution for 2y(y+3)=(y+4)(2y-1) equation: 



2y(y+3)=(y+4)(2y-1)

We move all terms to the left:

2y(y+3)-((y+4)(2y-1))=0

We multiply parentheses

2y^2+6y-((y+4)(2y-1))=0

We multiply parentheses ..

2y^2-((+2y^2-1y+8y-4))+6y=0



We calculate terms in parentheses: -((+2y^2-1y+8y-4)), so:

(+2y^2-1y+8y-4)

We get rid of parentheses

2y^2-1y+8y-4

We add all the numbers together, and all the variables

2y^2+7y-4

Back to the equation:

-(2y^2+7y-4)



We add all the numbers together, and all the variables

2y^2+6y-(2y^2+7y-4)=0

We get rid of parentheses

2y^2-2y^2+6y-7y+4=0

We add all the numbers together, and all the variables

-1y+4=0

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

-y=-4

y=-4/-1

y=+4

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