n(n-1)(n-2)(n-3)(n-4)=336

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Solution for n(n-1)(n-2)(n-3)(n-4)=336 equation: 



n(n-1)(n-2)(n-3)(n-4)=336

We move all terms to the left:

n(n-1)(n-2)(n-3)(n-4)-(336)=0

We multiply parentheses ..

n(+n^2-2n-1n+2)(n-3)(n-4)-336=0

We multiply parentheses ..

n(+n^2-2n-1n+2)(+n^2-4n-3n+12)-336=0

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

n(+n^2-2n-1n+2)(+n^2-4n-3n+12)=336

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