(2y+1)(y+3)-3y(y+2)=y(2-y)+10

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Solution for (2y+1)(y+3)-3y(y+2)=y(2-y)+10 equation: 



(2y+1)(y+3)-3y(y+2)=y(2-y)+10

We move all terms to the left:

(2y+1)(y+3)-3y(y+2)-(y(2-y)+10)=0

We add all the numbers together, and all the variables

(2y+1)(y+3)-3y(y+2)-(y(-1y+2)+10)=0

We multiply parentheses

-3y^2+(2y+1)(y+3)-6y-(y(-1y+2)+10)=0

We multiply parentheses ..

-3y^2+(+2y^2+6y+y+3)-6y-(y(-1y+2)+10)=0



We calculate terms in parentheses: -(y(-1y+2)+10), so:

y(-1y+2)+10

We multiply parentheses

-1y^2+2y+10

Back to the equation:

-(-1y^2+2y+10)



We get rid of parentheses

-3y^2+2y^2+1y^2+6y+y-2y-6y+3-10=0

We add all the numbers together, and all the variables

-1y-7=0

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

-y=7

y=7/-1

y=-7

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