(x+3)(x-4)=x(x-5)+2(x+6)

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Solution for (x+3)(x-4)=x(x-5)+2(x+6) equation: 



(x+3)(x-4)=x(x-5)+2(x+6)

We move all terms to the left:

(x+3)(x-4)-(x(x-5)+2(x+6))=0

We multiply parentheses ..

(+x^2-4x+3x-12)-(x(x-5)+2(x+6))=0



We calculate terms in parentheses: -(x(x-5)+2(x+6)), so:

x(x-5)+2(x+6)

We multiply parentheses

x^2-5x+2x+12

We add all the numbers together, and all the variables

x^2-3x+12

Back to the equation:

-(x^2-3x+12)



We get rid of parentheses

x^2-x^2-4x+3x+3x-12-12=0

We add all the numbers together, and all the variables

2x-24=0

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

2x=24

x=24/2

x=12

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