n(3+n)=n2+4n

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Solution for n(3+n)=n2+4n equation: 



n(3+n)=n2+4n

We move all terms to the left:

n(3+n)-(n2+4n)=0

We add all the numbers together, and all the variables

-(+n^2+4n)+n(n+3)=0

We multiply parentheses

-(+n^2+4n)+n^2+3n=0

We get rid of parentheses

-n^2+n^2-4n+3n=0

We add all the numbers together, and all the variables

-n=0

n=0/-1

n=0

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