How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have a solution.
My question is how I can find out how many solutions it has?

An idea to tackle this and similar questions using as small numbers and multiples of $\;101\;$ as possible (when possible, of course...). 
Observe that $\;5=-96\pmod {101}\;$ , and $\;96=2^5\cdot3\;$ , so we can try to deal with these apparently easier numbers. Since we have $\;96=16\cdot6\;$, we have:
$$4^2=16=2^4\;,\;\;101\cdot2=202-6=196=14^2$$
and thus we have: 
$$\;\sqrt{16}=4\;,\;\;\sqrt{-6}=14\implies \sqrt{16\cdot(-6)=-96}=4\cdot14=56$$
and thus also $\;\sqrt{-96}=-56=45\;$ , so $\;x^2=5\pmod{101}\implies x=\pm56=56,\,45\;$ 
Finally, your quadratic's solutions are ($\pmod{101}$ ):
$$x_{1,2}=\frac{-3\pm56}2=\begin{cases}\frac{98+56}2=77\\{}\\\frac{98-56}2=21\end{cases}$$
