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?

Since
$$
(x-49)^2-77\equiv x^2+3x+1\pmod{101}
$$
we are looking for solutions to $(x-49)^2\equiv77\pmod{101}$. You have verified that there is a root, so $77$ is a quadratic residue mod $101$, thus, there are two solutions for
$$
(x-49)^2\equiv77\pmod{101}
$$
Alternatively, working mod $101$, by squaring and multiplying
$$
77^2\equiv71\\
77^3\equiv13\\
77^6\equiv68\\
77^{12}\equiv79\\
77^{24}\equiv80\\
77^{25}\equiv100\\
77^{50}\equiv1
$$
Therefore, $77$ is a quadratic residue mod $101$. Thus, there are two solutions to $x^2+3x+1\pmod{101}$.
