(X-4)^2+6x=(x+3)(x-1)+6

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Solution for (X-4)^2+6x=(x+3)(x-1)+6 equation: 



(X-4)^2+6X=(X+3)(X-1)+6

We move all terms to the left:

(X-4)^2+6X-((X+3)(X-1)+6)=0

We add all the numbers together, and all the variables

6X+(X-4)^2-((X+3)(X-1)+6)=0

We multiply parentheses ..

-((+X^2-1X+3X-3)+6)+6X+(X-4)^2=0



We calculate terms in parentheses: -((+X^2-1X+3X-3)+6), so:

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

We get rid of parentheses

X^2-1X+3X-3+6

We add all the numbers together, and all the variables

X^2+2X+3

Back to the equation:

-(X^2+2X+3)



We add all the numbers together, and all the variables

6X-(X^2+2X+3)+(X-4)^2=0

We get rid of parentheses

-X^2+6X-2X+(X-4)^2-3=0

We add all the numbers together, and all the variables

-1X^2+4X+(X-4)^2-3=0

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

-1X^2+4X+(X-4)^2=3

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