Finding a tight bound for a recurrence relation
You just need to expant it (suppose $n$ is even) and using mathematical induction:
$$T(n) = n + (n-2) + T(n-4) = n + (n-2) + (n-4) + \cdots + 4 + 2 = $$
$$2 ( 1 + 2 + \cdots + \frac{n}{2}) = 2\frac{\frac{n}{2}(\frac{n}{2}+1)}{2} = \frac{n(n+2)}{4} = \Theta(n^2)$$