(x+7)^2=(x+3)(4x-2)

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Solution for (x+7)^2=(x+3)(4x-2) equation: 



(x+7)^2=(x+3)(4x-2)

We move all terms to the left:

(x+7)^2-((x+3)(4x-2))=0

We multiply parentheses ..

-((+4x^2-2x+12x-6))+(x+7)^2=0



We calculate terms in parentheses: -((+4x^2-2x+12x-6)), so:

(+4x^2-2x+12x-6)

We get rid of parentheses

4x^2-2x+12x-6

We add all the numbers together, and all the variables

4x^2+10x-6

Back to the equation:

-(4x^2+10x-6)



We get rid of parentheses

-4x^2-10x+(x+7)^2+6=0

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

-4x^2-10x+(x+7)^2=-6

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