6+(x-3)(x+4)=x(x+1)+12+3x

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Solution for 6+(x-3)(x+4)=x(x+1)+12+3x equation: 



6+(x-3)(x+4)=x(x+1)+12+3x

We move all terms to the left:

6+(x-3)(x+4)-(x(x+1)+12+3x)=0

We multiply parentheses ..

(+x^2+4x-3x-12)-(x(x+1)+12+3x)+6=0



We calculate terms in parentheses: -(x(x+1)+12+3x), so:

x(x+1)+12+3x

determiningTheFunctionDomain
x(x+1)+3x+12

We add all the numbers together, and all the variables

3x+x(x+1)+12

We multiply parentheses

x^2+3x+x+12

We add all the numbers together, and all the variables

x^2+4x+12

Back to the equation:

-(x^2+4x+12)



We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x-18=0

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

-3x=18

x=18/-3

x=-6

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