<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] One method is to form the square on the LHS : 
$$12 \cdot(3x^2+x+8) \equiv 0 \pmod{11}$$
$$36x^2+12x+1+95 \equiv 0 \pmod{11}$$
$$(6x+1)^2 \equiv -95 \equiv 4 \pmod{11}$$
This means that $$6x+1 \equiv \pm 2 \pmod{11}$$
When $6x+1 \equiv 2 \pmod{11}$ multiply by $2$ so :
$$12x \equiv 2\pmod{11}$$
$$x \equiv 2 \pmod{11}$$
Doing the same for the other leads to the other solution :
$$ x \equiv 5 \pmod{11}$$
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