Let $a,\,b,\,c$ be known integers and $p$ an odd prime number not dividing $a$.  The number of non-congruent roots of the quadratic congruence

 $\displaystyle ax^{2}\!+\!bx\!+\!c\;\equiv\;0\pmod{p}$ (1)

is

• two, if  $b^{2}\!-\!4ac$  is a quadratic residue modulo $p$;

• one, if  $b^{2}\!-\!4ac\equiv 0\pmod{p}$;

• zero, if  $b^{2}\!-\!4ac$  is a quadratic nonresidue modulo $p$.

Proof.  Since  $\gcd(p,\,4a)=1$,  multiplying (1) by $4a$ gives an equivalent (http://planetmath.org/Equivalent3) congruence

 $4a^{2}x^{2}\!+\!4abx\!+\!4ac\;\equiv\;0\pmod{p}$

which may furthermore be written as

 $(2ax\!+\!b)^{2}\;\equiv\;b^{2}\!-\!4ac\pmod{p}.$

Accordingly, one can obtain the the solution of the given congruence from the solution of the pair of congruences

 $\displaystyle\begin{cases}y^{2}\;\equiv\;b^{2}\!-\!4ac\pmod{p}\qquad\qquad(2)% \\ 2ax\!+\!b\;\equiv\;y\pmod{p}.\;\qquad\qquad(3)\\ \end{cases}$

Case 1:  $b^{2}\!-\!4ac$ is a quadratic residue$\pmod{p}$.  Then (2) has a root  $y=y_{0}\neq 0$,  and therefore also the second root  $y=-y_{0}$.  The roots  $y=\pm y_{0}$ are incongruent, because otherwise one had  $p\mid 2y_{0}$  and thus  $p\mid y_{0}\mid y_{0}^{2}\equiv b^{2}\!-\!4ac$  which is not possible in this case.
Case 2:  $b^{2}\!-\!4ac\equiv 0\pmod{p}$.  Now (2) implies that  $y\equiv 0\pmod{p}$,  whence the corresponding root $x_{0}$ of the linear congruence (3) does not allow other incongruent roots for (1).
Case 3:  $b^{2}\!-\!4ac$ is a quadratic nonresidue$\pmod{p}$.  The congruence (2) cannot have solutions; the same concerns thus also (1).

Example.  Solve the congruence

 $4x^{2}+6x-3\;\equiv\;0\pmod{43}.$

We have  $b^{2}\!-\!4ac=36+4\cdot 4\cdot 3=84\equiv-2\pmod{43}$  and the Legendre symbol

 $\left(\frac{-2}{43}\right)\;=\;\left(\frac{-1}{43}\right)\left(\frac{2}{43}% \right)\;=\;-1\cdot(-1)\;=\;1$

(see values of the Legendre symbol) says that $-2$ is a quadratic residue modulo 43.  The congruence corresponding (2) is

 $y^{2}\;\equiv\;-2\pmod{43},$

which is satisfied by  $y\equiv\pm 16\pmod{43}$ as one finds after a little experimenting.  Then we have the two linear congruences  $2\cdot 4x+6\equiv\pm 16\pmod{43}$,  i.e.

 $4x\;\equiv\;\pm 8-3\pmod{43}$

corresponding (3).  The first of them,  $4x\equiv 5\pmod{43}$,  is satisfied by  $x=12$  and the second,  $4x\equiv-11\pmod{43}$,  by  $x=8$.  Thus the solution of the given congruence is

 $x\;\equiv\;8\pmod{43}\quad\mbox{or}\quad x\;\equiv\;12\pmod{43}.$