(3x-1)^2-(2x+3)^2=5(x-2)+(x-1)(5x+2)

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Solution for (3x-1)^2-(2x+3)^2=5(x-2)+(x-1)(5x+2) equation: 



(3x-1)^2-(2x+3)^2=5(x-2)+(x-1)(5x+2)

We move all terms to the left:

(3x-1)^2-(2x+3)^2-(5(x-2)+(x-1)(5x+2))=0

We multiply parentheses ..

-(5(x-2)+(+5x^2+2x-5x-2))+(3x-1)^2-(2x+3)^2=0



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

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

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2+2x-12

Back to the equation:

-(5x^2+2x-12)



We get rid of parentheses

-5x^2-2x+(3x-1)^2-(2x+3)^2+12=0

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

-5x^2-2x+(3x-1)^2-(2x+3)^2=-12

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