(2x-1)^2-(x+1)^2=9x(x-3)=3x(3x-1)(x-1)=96

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Solution for (2x-1)^2-(x+1)^2=9x(x-3)=3x(3x-1)(x-1)=96 equation: 



(2x-1)^2-(x+1)^2=9x(x-3)=3x(3x-1)(x-1)=96

We move all terms to the left:

(2x-1)^2-(x+1)^2-(9x(x-3))=0



We calculate terms in parentheses: -(9x(x-3)), so:

9x(x-3)

We multiply parentheses

9x^2-27x

Back to the equation:

-(9x^2-27x)



We get rid of parentheses

-9x^2+(2x-1)^2-(x+1)^2+27x=0

We add all the numbers together, and all the variables

-9x^2+27x+(2x-1)^2-(x+1)^2=0

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