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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -(3(n+4)+55), so:

3(n+4)+55

We multiply parentheses

3n+12+55

We add all the numbers together, and all the variables

3n+67

Back to the equation:

-(3n+67)



We add all the numbers together, and all the variables

2n+2((n+2)+(n+3))-(3n+67)+1=0

We get rid of parentheses

2n+2((n+2)+(n+3))-3n-67+1=0

We add all the numbers together, and all the variables

-1n+2((n+2)+(n+3))-66=0

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

-1n+2((n+2)+(n+3))=66

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