(n-1/5)+(n-1/6)=(n^2-n/2)

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Solution for (n-1/5)+(n-1/6)=(n^2-n/2) equation: 



(n-1/5)+(n-1/6)=(n^2-n/2)

We move all terms to the left:

(n-1/5)+(n-1/6)-((n^2-n/2))=0

We add all the numbers together, and all the variables

(+n-1/5)+(+n-1/6)-((n^2-n/2))=0

We get rid of parentheses

n+n-((n^2-n/2))-1/5-1/6=0

We calculate fractions


n+n+(-((n^2-n*5*6)/()+()/()+()/()=0



We calculate terms in parentheses: +(-((n^2-n*5*6)/()+()/()+()/(), so:

-((n^2-n*5*6)/()+()/()+()/(

We can not solve this equation

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