. 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 $x= 31,102$.
The good old quadratic equation works just fine if $p\neq2$. If $a\not\equiv 0\pmod{p}$ and $b^2-4ac$ is a quadratic residue mod $p$, then the solutions to the quadratic congruence
$$ax^2+bx+c\equiv0\pmod{p},$$
are precisely
$$-\frac{b\pm\sqrt{b^2-4ac}}{2a}.$$
In this particular case we have $p=7$ and $a\equiv b\equiv1\pmod{7}$ and $c\equiv47\equiv5\pmod{7}$. Then
$$b^2-4ac\equiv2\equiv3^2\pmod{7},$$
so the congruence has the two solutions
$$-\frac{1+3}{2}=5\qquad\text{ and }\qquad-\frac{1-3}{2}=1.$$
On the other hand, if you want to solve it purely by inspection, note that there are only $7$ possible solutions to check. Clearly $x=0$ is not a solution, and plugging in $x=1$ yields the first solution. The product of the solutions is congruent to $47\equiv5\pmod{7}$, so the other solution is $x=5$.
Best Answer
Theorem: a polynomial of degree $n$ has at most $n$ zeroes $\!\bmod p$.
Proof: $ax\equiv b\pmod {\!p}$ has one solution. Assume that any polynomial of degree $n-1$ has at most $n-1$ roots. Consider an arbitrary polynomial $f$ of degree $n$. If it has no zeroes, we're done. Otherwise (let the root be $a$) we can write $f$ as $f(x)=(x-a)g(x)$, where $g$ is a polynomial of degree $n-1$. $f$ then must have at most $1+(n-1)=n$ zeroes. $\ \ \ \square$
Not true for coprime moduli. E.g., $x^2\equiv 1\pmod {\!12}\iff x\equiv \{1,5,7,11\}$.
CRT implies every polynomial of degree $n$ has at most $n^k$ roots mod $p_1p_2\cdots p_k$, where $p_i$ are distinct primes, since any combination of residues mod $p_i$ gives a unique residue mod $p_1p_2\cdots p_k$ and vice versa, and there are $\underbrace{n\cdot n\cdots n}_{k\text{ }n\text{'s}}=n^k$ such combinations.