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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((y-3)(y-2)), so:

(y-3)(y-2)

We multiply parentheses ..

(+y^2-2y-3y+6)

We get rid of parentheses

y^2-2y-3y+6

We add all the numbers together, and all the variables

y^2-5y+6

Back to the equation:

-(y^2-5y+6)


We get rid of parentheses

y^2-y^2+2y-1y+5y-2-6=0

We add all the numbers together, and all the variables

6y-8=0

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

6y=8

y=8/6

y=1+1/3