. To find solutions of $x^2 - 6x - 13 \equiv 0 \mod 127$, we must write $x^2-6x - 13$ as a square of a linear term, plus a constant. It's seen easily that $x^2 - 6x - 13 = (x-3)^2 - 22$. Hence, the congruence is equivalent to $(x-3)^2 \equiv 22 \mod 127$.
Now, we must check if $22$ is a quadratic residue mod $127$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused.
$$
\binom{22}{127} = \color{green}{\binom{11}{127}}\color{red}{\binom{2}{127}} = \color{green}{-\binom{127}{11}} \times \color{red}1
$$
where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $\binom{2}{p}$ is well known by the remainders which $p$ leaves when divided by $8$.
Now, we can do:
$$
-\binom{127}{11} = -\binom{6}{11} = -\color{green}{\binom{2}{11}}\color{red}{ \binom{3}{11}} = -\color{green}{(-1)}\color{red}{(1)} = 1
$$
Again, the colored terms on the LHS and RHS are equal because the quantities $\binom 2p$ and $\binom 3p$ are well known.
Since we have obtained that the Legendre symbol is $1$, this implies the existence of a solution, and therefore two.
The question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $(x-3)^2 \equiv 22 \mod 127$ is that we basically only need to look for squares of the form $127k + 22$ to find a solution, rather than having to substitute values of $x$ into the expression $x^2-6x-13$ each time.
A brute force : the series $127k + 22 $ goes like : $22,149,276,403,530,657,\color{blue}{784},...$
lo and behold, $784 = 28^2$, hence this gives $x = 31$. Now, note that the congruence actually has two solutions, one given by $127 - 28 = 99$. You can check that $9801 = 99^2 = 22 + 127 \times 77$.
Hence, we get two solutions of $x$, namely $x= 31,102$.
Best Answer
Hint $ $ Let $\,f(x) = x^2-1\,$ below and exploit the highlighted $\rm\color{#c00}{multiplicativity}$ of number of roots.
Remark $\ $ If $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ & mod $\,n.\,$ By CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn,\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{mn}&\overset{\rm \large CRT\!\!}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_{\color{}{\large k}}\pmod{\! m},\phantom{I^{I^{I^I}}}\ \ \color{#c00}k\ \ \rm roots\\x\equiv s_1,\ldots,s_{\large\color{}{\ell}}\pmod n,\ \ \ \ \ \ \ \ \ \,\color{#c00}{\ell}\ \ \rm roots\end{array}\\[.3em] &\iff& \left\{ \begin{array}{}x\equiv r_{\large i}\pmod{\! m}\\x\equiv s_{\large j}\pmod n\end{array} \right\}\ \ \ \ {\rm for}\ \ \begin{array}{}1 \le i \le k\\ 1\le j\le\ell\end{array}\\ &\overset{\rm\large CRT\!}\iff& \left\{ x\equiv t_{\,\large i\, j}\!\!\!\!\pmod{\!mn} \right\}\,\ \ \underbrace{{\rm for}\ \ 1 \le i\le k,\,\ 1\le j\le\ell}_{\Large \color{#c00}{k\,\cdot\, \ell}\ \,\rm roots\ \ \ \ \ \ \ \ \ \ }\\ \end{eqnarray}\qquad\qquad$$