I need to find the primes for which the congruential $x^2-4x-3=0$ (mod $p$). First I checked for $p=2$ and $x=1$ is a solution. Assuming $p>2$, my approach was that the discriminant $=28$ needs to be a quadratic residue modulo $p$.
Using Legendre's symbol, $(28/p)=(7/p)(4/p)=(7/p)$ (as $4$ is a quadratic residue). Now I need to find for which primes is $7$ a quadratic residue. I sense that I need to use the Quadratic Reciprocity to continue, but when I try I get stuck with splitting to separate cases.
I couldn't find a way or an algorithm online to solve such a question other than equations involving only $(-1/p)$, $(2/p)$, $(3/p)$.
Could someone show me how to continue from the spot I got stuck?