(n-2)*(n-3)*(n-4)*(n-5)=4

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Solution for (n-2)*(n-3)*(n-4)*(n-5)=4 equation: 



(n-2)(n-3)(n-4)(n-5)=4

We move all terms to the left:

(n-2)(n-3)(n-4)(n-5)-(4)=0

We multiply parentheses ..

(+n^2-3n-2n+6)(n-4)(n-5)-4=0

We multiply parentheses ..

(+n^2-3n-2n+6)(+n^2-5n-4n+20)-4=0

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

(+n^2-3n-2n+6)(+n^2-5n-4n+20)=4

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