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

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



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

We move all terms to the left:

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

We multiply parentheses ..

-((+2x^2+10x-3x-15)-3)+(2x+1)^2=0



We calculate terms in parentheses: -((+2x^2+10x-3x-15)-3), so:

(+2x^2+10x-3x-15)-3

We get rid of parentheses

2x^2+10x-3x-15-3

We add all the numbers together, and all the variables

2x^2+7x-18

Back to the equation:

-(2x^2+7x-18)



We get rid of parentheses

-2x^2-7x+(2x+1)^2+18=0

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

-2x^2-7x+(2x+1)^2=-18

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