Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$ Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$
I think it has no solutions but I am not sure how to show this.

We have $4x^2+6x+1 = (2x+3/2)^2-5/4$. Hence,
\begin{align}
4x^2+6x+1 \equiv 0 \pmod{13} \implies (2x+3/2)^2-5/4 \equiv 0 \pmod{13}
\end{align}
This simplifies into
\begin{align}
(4x+3)^2 \equiv 5 \pmod{13}
\end{align}
This has no solutions, since
$$5^{(13-1)/2}\pmod{13} \equiv 5^6\pmod{13} \equiv (25)^3 \pmod{13} \equiv-1\pmod{13}$$
