Modular Quadratic Equation I'm trying to solve that equation:
$x^2-3x-5\equiv0\pmod{343}$
I've completed the square as follows:
$x^2-3x-5 \equiv x^2+340x-5\equiv(x+170)^2-170^2-5\pmod{343}\\
(x+170)^2 \equiv 93\pmod{343}\\
y^2 \equiv 93 \pmod{343}$
But I have no idea how to move on. How can I use the fact that $343=7^3$?

HINT:
Start with $x^2-3x-5\equiv0\pmod7\iff0\equiv x^2-3x-5+7=(x-1)(x-2)$
If $x-1\equiv0\pmod7, x=7a+1$ where $a$ is any integer
Now $x^2-3x-5=(7a+1)^2-3(7a+1)-5=49a^2-7a-7$
which $\equiv0\pmod{7^2}\iff7|(a+1)\implies a\equiv-1\pmod7, a=7b-1$ where $b$ any integer
$\implies x=7a+1=7(7b-1)+1=49b-6$
Now $x^2-3x-5=(49b-6)^2-3(49b-6)-5\equiv49(-15b+1)\pmod{7^3}$
which will be $\equiv0\iff-15b+1\equiv0\pmod7\iff b\equiv1$
