Solve Quadratic Congruence Equation How to solve $3x^2 - 5x + 5 \equiv 0 \pmod 7$? In general, how to approach this kind of problem? Any help is appreciated.

If $a,b,c,x,n\in\mathbb Z$, $n\ge 3$, $\gcd(2a,n)=1$, then
$$ax^2+bx+c\equiv 0\pmod{n}$$
$$\stackrel{\cdot 4a}\iff (2ax+b)^2\equiv b^2-4ac\pmod{n}$$
You're left with finding the modular square roots of $b^2-4ac$ mod $n$.
First, you may want to find a Legendre symbol, use Quadratic reciprocity.
Here you're solving $3x^2 - 5x + 5 \equiv 0 \pmod 7$.
In this case, you're left with finding the modular square roots of $(-5)^2-4(3)(5)\equiv -35\equiv 0$ mod $7$, which is just $0$.
$2(3)x-5\equiv 0\pmod{7}$, i.e. $6x\equiv 5\pmod{7}$,
i.e. $-x\equiv 5\pmod{7}$. Multiply both sides by $-1$ (here $\gcd(-1,7)=1$).
$\iff x\equiv -5\equiv 2\pmod{7}$.
