(X+2)(x+3)=(x-2)(x-4)+7

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Solution for (X+2)(x+3)=(x-2)(x-4)+7 equation: 



(X+2)(X+3)=(X-2)(X-4)+7

We move all terms to the left:

(X+2)(X+3)-((X-2)(X-4)+7)=0

We multiply parentheses ..

(+X^2+3X+2X+6)-((X-2)(X-4)+7)=0



We calculate terms in parentheses: -((X-2)(X-4)+7), so:

(X-2)(X-4)+7

We multiply parentheses ..

(+X^2-4X-2X+8)+7

We get rid of parentheses

X^2-4X-2X+8+7

We add all the numbers together, and all the variables

X^2-6X+15

Back to the equation:

-(X^2-6X+15)



We get rid of parentheses

X^2-X^2+3X+2X+6X+6-15=0

We add all the numbers together, and all the variables

11X-9=0

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

11X=9

X=9/11

X=9/11

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