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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+x^2+5x-3x-15))+(2x+3)^2=0



We calculate terms in parentheses: -((+x^2+5x-3x-15)), so:

(+x^2+5x-3x-15)

We get rid of parentheses

x^2+5x-3x-15

We add all the numbers together, and all the variables

x^2+2x-15

Back to the equation:

-(x^2+2x-15)



We get rid of parentheses

-x^2-2x+(2x+3)^2+15=0

We add all the numbers together, and all the variables

-1x^2-2x+(2x+3)^2+15=0

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

-1x^2-2x+(2x+3)^2=-15

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