Solutions of $x^2-6x-13 \equiv 0 \pmod{127}$ I started learning number theory, specifically polynomial congruences, and need help with the following exercise. Here it is:

Does the congruence $x^2-6x-13 \equiv 0 \pmod{127}$ has solutions? 

I tried to follow the method for solving general quadratic congruence but I didn't get really far. Here's what I've done so far:
Since $(4, 127) = 1$, we may complete the square by multiplying by $4$ without having to change the modulus in order to get the following equivalent congruence
$$(2x-6)^2 \equiv 36 - 4(-13) \pmod{127} \iff (2x-6)^2 \equiv 88 \pmod{127}.$$
If I'm heading in the right direction then I don't know how to continue from here. I suspect there is another method for solving this problem since I didn't make use of the fact that $127$ is prime. Also, the problem doesn't require to actually find the solutions but only determine if there are solutions. Possibly we can avoid computations and make use of some theorem/lemma to find if the congruence has solutions or not. 
 A: . 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 $x= 31,102$.
A: Using the Berlekamp Algorithm for factoring a polynomial over $\mathbb{F}_{127}$ we immediately obtain that
$$
x^2-6x-13=(x+25)(x+96).
$$
So we have two solutions.
A: I think you can try $$x=31.$$ It is valid.
Finally we have
$$x^2-6x-13\equiv(x-31)(x-102) \pmod{127}.$$
A: This equation has roots modulo $127$ if and only if the reduced discriminant $\Delta'=9+13$ is a square modulo $127$. We'll use the laws of quadratic reciprocity.
$$\biggl(\frac{22}{127}\biggr)=\biggl(\frac{2}{127}\biggr)\biggl(\frac{11}{127}\biggr)=\biggl(\frac{11}{127}\biggr)\quad\text{since }127\equiv -1\mod8\quad (2^{\mathit{nd}}\textit{supplementary law}). $$
Now we have
\begin{align}\biggl(\frac{11}{127}\biggr)&=\biggl(\frac{127}{11}\biggr)\bigl(-1\bigr)^{\tfrac{10\cdot126}4}=-\biggl(\frac{6}{11}\biggr)=-\biggl(\frac{2}{11}\biggr)\biggl(\frac{3}{11}\biggr)\\[1ex]
&=+\biggl(\frac{3}{11}\biggr)=\biggl(\frac{11}{3}\biggr)\bigl(-1\bigr)^{\tfrac{2\cdot10}4}=-\biggl(\frac{2}{3}\biggr)=+1.
\end{align}
Thus the equation has roots.
