(2y+3)(y-4)=(y-1)(2y+5)

Solution for (2y+3)(y-4)=(y-1)(2y+5) equation:

(2y+3)(y-4)=(y-1)(2y+5)

We move all terms to the left:

(2y+3)(y-4)-((y-1)(2y+5))=0

We multiply parentheses ..

(+2y^2-8y+3y-12)-((y-1)(2y+5))=0


We calculate terms in parentheses: -((y-1)(2y+5)), so:

(y-1)(2y+5)

We multiply parentheses ..

(+2y^2+5y-2y-5)

We get rid of parentheses

2y^2+5y-2y-5

We add all the numbers together, and all the variables

2y^2+3y-5

Back to the equation:

-(2y^2+3y-5)


We get rid of parentheses

2y^2-2y^2-8y+3y-3y-12+5=0

We add all the numbers together, and all the variables

-8y-7=0

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

-8y=7

y=7/-8

y=-7/8