Gauss won't help

Gauss does help... he was the one who studied this very expression (for arbitrary primes)!

We will find $\sum_{n=1}^{p-2}\left(\frac{n(n+1)}{p}\right)$

for any odd prime number $p$ .

For any $0<n<p-1$ there exists an $0<a(n)<p-1$ such that $n\times a(n)\equiv 1 \pmod{p}$ : the multiplicative inverse.

Now

$\sum_{n=1}^{p-2}\left(\frac{n(n+1)}{p}\right)=\sum_{n=1}^{p-2}\left(\frac{n(n+n\times a(n))}{p}\right)=\sum_{n=1}^{p-2}\left(\frac{n^2(1+a(n))}{p}\right)=\sum_{a=1}^{p-2}\left(\frac{1+a}{p}\right)=\boxed{-1}$

The last equation holds since $\left(\frac{1}{p}\right)=1$ is missing from the sum.