(3x-9)^2=(2x+2)(2x+29)

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Solution for (3x-9)^2=(2x+2)(2x+29) equation: 



(3x-9)^2=(2x+2)(2x+29)

We move all terms to the left:

(3x-9)^2-((2x+2)(2x+29))=0

We multiply parentheses ..

-((+4x^2+58x+4x+58))+(3x-9)^2=0



We calculate terms in parentheses: -((+4x^2+58x+4x+58)), so:

(+4x^2+58x+4x+58)

We get rid of parentheses

4x^2+58x+4x+58

We add all the numbers together, and all the variables

4x^2+62x+58

Back to the equation:

-(4x^2+62x+58)



We get rid of parentheses

-4x^2-62x+(3x-9)^2-58=0

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

-4x^2-62x+(3x-9)^2=58

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