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

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



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

We move all terms to the left:

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

We multiply parentheses ..

-((+3x^2-4x+6x-8))+(2x-1)^2-(x-3)^2=0



We calculate terms in parentheses: -((+3x^2-4x+6x-8)), so:

(+3x^2-4x+6x-8)

We get rid of parentheses

3x^2-4x+6x-8

We add all the numbers together, and all the variables

3x^2+2x-8

Back to the equation:

-(3x^2+2x-8)



We get rid of parentheses

-3x^2-2x+(2x-1)^2-(x-3)^2+8=0

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

-3x^2-2x+(2x-1)^2-(x-3)^2=-8

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