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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

2y^2+(-1y+8)-((2y+3)(y-4))=0

We get rid of parentheses

2y^2-1y-((2y+3)(y-4))+8=0

We multiply parentheses ..

2y^2-((+2y^2-8y+3y-12))-1y+8=0


We calculate terms in parentheses: -((+2y^2-8y+3y-12)), so:

(+2y^2-8y+3y-12)

We get rid of parentheses

2y^2-8y+3y-12

We add all the numbers together, and all the variables

2y^2-5y-12

Back to the equation:

-(2y^2-5y-12)


We add all the numbers together, and all the variables

2y^2-1y-(2y^2-5y-12)+8=0

We get rid of parentheses

2y^2-2y^2-1y+5y+12+8=0

We add all the numbers together, and all the variables

4y+20=0

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

4y=-20

y=-20/4

y=-5