Combination Problem: Solving for $n$ such that $C(n, 2) = 10$

Question

Solve for n ,

$C(n, 2) = 10$

What I have done so far:

$\frac{n!}{2!(n-2)!} = 10$

$\frac{n!}{2!(n-2)(n-1)n!} =10$

$\frac{1}{2n^2 - 6n-4} = 10$

$ 1 = 20n^2-6n-40$

I think I did something wrong along the way

Answer 1

$C(n,2)=\frac{n!}{(n-2)!2!}=\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n^2-n}2$, so you need to solve $n^2-n=20$.