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

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



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

We move all terms to the left:

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

We multiply parentheses ..

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



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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-7x-15

Back to the equation:

-(2x^2-7x-15)



We get rid of parentheses

-2x^2+7x+2(x+2)^2+15+1=0

We add all the numbers together, and all the variables

-2x^2+7x+2(x+2)^2+16=0

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

-2x^2+7x+2(x+2)^2=-16

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