(x+1)(x+2)+(x+2)(x+3)=2(x^2+5x+6)

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Solution for (x+1)(x+2)+(x+2)(x+3)=2(x^2+5x+6) equation: 



(x+1)(x+2)+(x+2)(x+3)=2(x^2+5x+6)

We move all terms to the left:

(x+1)(x+2)+(x+2)(x+3)-(2(x^2+5x+6))=0

We multiply parentheses ..

(+x^2+2x+x+2)+(x+2)(x+3)-(2(x^2+5x+6))=0



We calculate terms in parentheses: -(2(x^2+5x+6)), so:

2(x^2+5x+6)

We multiply parentheses

2x^2+10x+12

Back to the equation:

-(2x^2+10x+12)



We get rid of parentheses

x^2-2x^2+2x+x+(x+2)(x+3)-10x+2-12=0

We multiply parentheses ..

x^2-2x^2+(+x^2+3x+2x+6)+2x+x-10x+2-12=0

We add all the numbers together, and all the variables

-1x^2+(+x^2+3x+2x+6)-7x-10=0

We get rid of parentheses

-1x^2+x^2+3x+2x-7x+6-10=0

We add all the numbers together, and all the variables

-2x-4=0

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

-2x=4

x=4/-2

x=-2

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