n+(n+1)+2((n+2)+(n+3)+55)=3(n+4)

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Solution for n+(n+1)+2((n+2)+(n+3)+55)=3(n+4) equation: 



n+(n+1)+2((n+2)+(n+3)+55)=3(n+4)

We move all terms to the left:

n+(n+1)+2((n+2)+(n+3)+55)-(3(n+4))=0

We get rid of parentheses

n+n+2((n+2)+(n+3)+55)-(3(n+4))+1=0



We calculate terms in parentheses: -(3(n+4)), so:

3(n+4)

We multiply parentheses

3n+12

Back to the equation:

-(3n+12)



We add all the numbers together, and all the variables

2n+2((n+2)+(n+3)+55)-(3n+12)+1=0

We get rid of parentheses

2n+2((n+2)+(n+3)+55)-3n-12+1=0

We add all the numbers together, and all the variables

-1n+2((n+2)+(n+3)+55)-11=0

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

-1n+2((n+2)+(n+3)+55)=11

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