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

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



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

We move all terms to the left:

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

We multiply parentheses ..

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



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

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

We multiply parentheses ..

(+x^2-2x-5x+10)+2

We get rid of parentheses

x^2-2x-5x+10+2

We add all the numbers together, and all the variables

x^2-7x+12

Back to the equation:

-(x^2-7x+12)



We get rid of parentheses

x^2-x^2+4x-3x+7x-12-12=0

We add all the numbers together, and all the variables

8x-24=0

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

8x=24

x=24/8

x=3

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