I was reading the IB mathematics book and I came across an equation related to factorial and ratios. The equation is 
$$\frac{(n+1)!}{(n-2)!} = 990$$

so these types of equations generally lead to a quadratic of some sort however this one leads to $n^3 - n = 990$. I have no idea how to solve this, the answer is $10$ but I have no clue how I end up with that. I can obviously guess that from looking at the equation, but what is the correct way to do it?

\*it is stated in the answer sheet that quadratic is the way to go but I can't really figure out how

entire process of what i do is : 

$$(n+1)n(n-1)(n-2)! / (n-2)! = 990 ~> (n+1)n(n-1) = 990$$

$$(n^2+n)(n-1) = 990 ~> n^3 + n^2 - n^2 - n = 990 ~> n^3 - n = 990$$

tried doing: $n (n^2-1) = 900$ ~> $n^2 - 1 = 900/n$ ~> $n^2 - 900/n -1$ but didn't really help me.

Thank you for your response.