(n-6)(3n+1)=(3n)(n-5)

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Solution for (n-6)(3n+1)=(3n)(n-5) equation: 



(n-6)(3n+1)=(3n)(n-5)

We move all terms to the left:

(n-6)(3n+1)-((3n)(n-5))=0

We multiply parentheses ..

(+3n^2+n-18n-6)-(3n(n-5))=0



We calculate terms in parentheses: -(3n(n-5)), so:

3n(n-5)

We multiply parentheses

3n^2-15n

Back to the equation:

-(3n^2-15n)



We get rid of parentheses

3n^2-3n^2+n-18n+15n-6=0

We add all the numbers together, and all the variables

-2n-6=0

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

-2n=6

n=6/-2

n=-3

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