<s>[INST] Solve the congruence $3x^2+x+8\equiv 0 \pmod{11}$ How to find the solutions of this congruence? $$3x^2+x+8\equiv 0 \pmod{11}$$ 
I need to find the inverse of $3$, and there I have a problem.
 [/INST] In general, if $p$ is an odd prime and $p\nmid a$, then $$ax^2+bx+c\equiv 0\pmod{p}\iff 4a\left(ax^2+bx+c\right)\equiv 0\pmod{p}$$
$$\iff (2ax+b)^2\equiv b^2-4ac\pmod{p}$$
At this point, you're just left with finding the square roots of $b^2-4ac$ mod $p$.

In your case, $$3x^2+x+8\equiv 0\pmod{11}\iff (6x+1)^2\equiv 4\pmod{11}$$
In this case, the square roots of $4$ mod $p$ are simply $\pm 2$, so:
$$\iff 6x+1\equiv \pm 2\pmod{11}$$
$$\iff x\equiv \left\{\frac{2-1}{6},\frac{-2-1}{6}\right\}\equiv \left\{\frac{1}{6},\frac{-3}{6}\right\}\pmod{11}$$
$$\iff x\equiv \left\{\frac{12}{6}, -\frac{1}{2}\right\}\equiv \left\{2,-\frac{12}{2}\right\}\equiv \{2,-6\}\equiv \{2,5\}\pmod{11}$$

I'll explain about the general solution.
As I said, you're just left with finding the square roots of $b^2-4ac$ mod $p$ (after that it's simple).
So you're left with solving $X^2\equiv A\pmod{p}$ to find $X$,
where $X\equiv 2ax+b\pmod{p}$ and $A\equiv b^2-4ac\pmod{p}$.
A solution exists if and only if $\left(\frac{A}{p}\right)\in\{0,1\}$. This denotes the Legendre Symbol. Finding it might require Quadratic Reciprocity.
If a solution exists, then see this paper and Wikipedia for how to find the solutions.
If $p\equiv 3\pmod{4}$, then $X\equiv \pm A^{\frac{p+1}{4}}\pmod{p}$.
If $p\equiv 5\pmod{8}$, then either $X\equiv \pm A^{\frac{p+3}{8}}\pmod{p}$ or $X\equiv \pm 2^{\frac{p-1}{4}}A^{\frac{p+3}{8}}\pmod{p}$.
If $p\equiv 1\pmod{8}$, then it's complicated. You may need to apply either Cipolla's Algorithm or Tonelli-Shanks algorithm, or see the paper I linked.
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