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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(y-1)(y-1)+2

We multiply parentheses ..

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

We get rid of parentheses

y^2-1y-1y+1+2

We add all the numbers together, and all the variables

y^2-2y+3

Back to the equation:

-(y^2-2y+3)


We multiply parentheses

2y^2-4y-4y-(y-5)(y+3)-(y^2-2y+3)+8=0

We get rid of parentheses

2y^2-y^2-4y-4y-(y-5)(y+3)+2y-3+8=0

We multiply parentheses ..

2y^2-y^2-(+y^2+3y-5y-15)-4y-4y+2y-3+8=0

We add all the numbers together, and all the variables

y^2-(+y^2+3y-5y-15)-6y+5=0

We get rid of parentheses

y^2-y^2-3y+5y-6y+15+5=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