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

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



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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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



We calculate terms in parentheses: -((+3x^2+6x+3x+6)), so:

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

We get rid of parentheses

3x^2+6x+3x+6

We add all the numbers together, and all the variables

3x^2+9x+6

Back to the equation:

-(3x^2+9x+6)



We add all the numbers together, and all the variables

3x^2+12x-(3x^2+9x+6)=0

We get rid of parentheses

3x^2-3x^2+12x-9x-6=0

We add all the numbers together, and all the variables

3x-6=0

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

3x=6

x=6/3

x=2

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