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

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



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

We move all terms to the left:

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

We multiply parentheses ..

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



We calculate terms in parentheses: -(x^2+(x+1)(x+6)+x), so:

x^2+(x+1)(x+6)+x

We add all the numbers together, and all the variables

x^2+x+(x+1)(x+6)

We multiply parentheses ..

x^2+(+x^2+6x+x+6)+x

We get rid of parentheses

x^2+x^2+6x+x+x+6

We add all the numbers together, and all the variables

2x^2+8x+6

Back to the equation:

-(2x^2+8x+6)



We get rid of parentheses

2x^2-2x^2+6x+2x-8x+6-6=0

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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