(X+3)^2+(x-3)^2=(x+6)(x-1)+20

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Solution for (X+3)^2+(x-3)^2=(x+6)(x-1)+20 equation: 



(X+3)^2+(X-3)^2=(X+6)(X-1)+20

We move all terms to the left:

(X+3)^2+(X-3)^2-((X+6)(X-1)+20)=0

We multiply parentheses ..

-((+X^2-1X+6X-6)+20)+(X+3)^2+(X-3)^2=0



We calculate terms in parentheses: -((+X^2-1X+6X-6)+20), so:

(+X^2-1X+6X-6)+20

We get rid of parentheses

X^2-1X+6X-6+20

We add all the numbers together, and all the variables

X^2+5X+14

Back to the equation:

-(X^2+5X+14)



We get rid of parentheses

-X^2-5X+(X+3)^2+(X-3)^2-14=0

We add all the numbers together, and all the variables

-1X^2-5X+(X+3)^2+(X-3)^2-14=0

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

-1X^2-5X+(X+3)^2+(X-3)^2=14

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