n(n+1)(n+2)(n+3)=2018

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Solution for n(n+1)(n+2)(n+3)=2018 equation: 



n(n+1)(n+2)(n+3)=2018

We move all terms to the left:

n(n+1)(n+2)(n+3)-(2018)=0

We multiply parentheses ..

n(+n^2+2n+n+2)(n+3)-2018=0

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

n(+n^2+2n+n+2)(n+3)=2018

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