(x+3)(x2+2x+6)=-(x+3)(x-4)

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Solution for (x+3)(x2+2x+6)=-(x+3)(x-4) equation: 



(x+3)(x2+2x+6)=-(x+3)(x-4)

We move all terms to the left:

(x+3)(x2+2x+6)-(-(x+3)(x-4))=0

We add all the numbers together, and all the variables

(x+3)(+x^2+2x+6)-(-(x+3)(x-4))=0

We multiply parentheses ..

(x+3)(+x^2+2x+6)-(-(+x^2-4x+3x-12))=0



We calculate terms in parentheses: -(-(+x^2-4x+3x-12)), so:

-(+x^2-4x+3x-12)

We get rid of parentheses

-x^2+4x-3x+12

We add all the numbers together, and all the variables

-1x^2+x+12

Back to the equation:

-(-1x^2+x+12)



We get rid of parentheses

(x+3)(+x^2+2x+6)+1x^2-x-12=0

We add all the numbers together, and all the variables

x^2+(x+3)(+x^2+2x+6)-1x-12=0

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

x^2+(x+3)(+x^2+2x+6)-1x=12

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