N(x)=100(3x+20)

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Solution for N(x)=100(3x+20) equation: 



(N)=100(3N+20)

We move all terms to the left:

(N)-(100(3N+20))=0



We calculate terms in parentheses: -(100(3N+20)), so:

100(3N+20)

We multiply parentheses

300N+2000

Back to the equation:

-(300N+2000)



We get rid of parentheses

N-300N-2000=0

We add all the numbers together, and all the variables

-299N-2000=0

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

-299N=2000

N=2000/-299

N=-6+206/299

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