(n+2)(2-n)n=(n-2)(2-n)5-(n-2)(n+2)(n-1)

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



(n+2)(2-n)n=(n-2)(2-n)5-(n-2)(n+2)(n-1)

We move all terms to the left:

(n+2)(2-n)n-((n-2)(2-n)5-(n-2)(n+2)(n-1))=0

We add all the numbers together, and all the variables

(n+2)(-1n+2)n-((n-2)(-1n+2)5-(n-2)(n+2)(n-1))=0

We multiply parentheses ..

(-1n^2+2n-2n+4)n-((n-2)(-1n+2)5-(n-2)(n+2)(n-1))=0



We calculate terms in parentheses: -((n-2)(-1n+2)5-(n-2)(n+2)(n-1)), so:

(n-2)(-1n+2)5-(n-2)(n+2)(n-1)

We multiply parentheses ..

(-1n^2+2n+2n-4)5-(n-2)(n+2)(n-1)

We multiply parentheses

-5n^2+10n+10n-(n-2)(n+2)(n-1)-20

We multiply parentheses ..

-5n^2-(+n^2+2n-2n-4)(n-1)+10n+10n-20

We add all the numbers together, and all the variables

-5n^2-(+n^2+2n-2n-4)(n-1)+20n-20

Back to the equation:

-(-5n^2-(+n^2+2n-2n-4)(n-1)+20n-20)



We multiply parentheses

-1n^3+2n^2-2n^2-(-5n^2-(+n^2+2n-2n-4)(n-1)+20n-20)+4n=0



We calculate terms in parentheses: -(-5n^2-(+n^2+2n-2n-4)(n-1)+20n-20), so:

-5n^2-(+n^2+2n-2n-4)(n-1)+20n-20



We do not support enpression: n^3

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