n+(n+1)+(n+2)+(n+3)=138

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Solution for n+(n+1)+(n+2)+(n+3)=138 equation: 



n+(n+1)+(n+2)+(n+3)=138

We move all terms to the left:

n+(n+1)+(n+2)+(n+3)-(138)=0

We get rid of parentheses

n+n+n+n+1+2+3-138=0

We add all the numbers together, and all the variables

4n-132=0

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

4n=132

n=132/4

n=33

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