(x-3)(2x-3)-6x=(x-4)(2x+4)+12

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Solution for (x-3)(2x-3)-6x=(x-4)(2x+4)+12 equation: 



(x-3)(2x-3)-6x=(x-4)(2x+4)+12

We move all terms to the left:

(x-3)(2x-3)-6x-((x-4)(2x+4)+12)=0

We add all the numbers together, and all the variables

-6x+(x-3)(2x-3)-((x-4)(2x+4)+12)=0

We multiply parentheses ..

(+2x^2-3x-6x+9)-6x-((x-4)(2x+4)+12)=0



We calculate terms in parentheses: -((x-4)(2x+4)+12), so:

(x-4)(2x+4)+12

We multiply parentheses ..

(+2x^2+4x-8x-16)+12

We get rid of parentheses

2x^2+4x-8x-16+12

We add all the numbers together, and all the variables

2x^2-4x-4

Back to the equation:

-(2x^2-4x-4)



We get rid of parentheses

2x^2-2x^2-3x-6x-6x+4x+9+4=0

We add all the numbers together, and all the variables

-11x+13=0

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

-11x=-13

x=-13/-11

x=1+2/11

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