(y-4)(y-4)-7(2-y)=(y+5)(y-2)

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Solution for (y-4)(y-4)-7(2-y)=(y+5)(y-2) equation: 



(y-4)(y-4)-7(2-y)=(y+5)(y-2)

We move all terms to the left:

(y-4)(y-4)-7(2-y)-((y+5)(y-2))=0

We add all the numbers together, and all the variables

(y-4)(y-4)-7(-1y+2)-((y+5)(y-2))=0

We multiply parentheses

(y-4)(y-4)+7y-((y+5)(y-2))-14=0

We multiply parentheses ..

(+y^2-4y-4y+16)+7y-((y+5)(y-2))-14=0



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

(y+5)(y-2)

We multiply parentheses ..

(+y^2-2y+5y-10)

We get rid of parentheses

y^2-2y+5y-10

We add all the numbers together, and all the variables

y^2+3y-10

Back to the equation:

-(y^2+3y-10)



We get rid of parentheses

y^2-y^2-4y-4y+7y-3y+16+10-14=0

We add all the numbers together, and all the variables

-4y+12=0

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

-4y=-12

y=-12/-4

y=+3

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