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

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



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

We move all terms to the left:

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

We multiply parentheses

(x+3)^2-2x-((x+1)(x+2)-5)-10=0

We multiply parentheses ..

-((+x^2+2x+x+2)-5)+(x+3)^2-2x-10=0



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

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

We get rid of parentheses

x^2+2x+x+2-5

We add all the numbers together, and all the variables

x^2+3x-3

Back to the equation:

-(x^2+3x-3)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-5x+(x+3)^2-7=0

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

-1x^2-5x+(x+3)^2=7

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