Find the solution of the $x^2+ 2x +3=0$ mod 198 Find the solution of the  $x^2+2x+3 \equiv0\mod{198}$
i have no idea for this problem i have small hint to we going consider $x^2+2x+3 \equiv0\mod{12}$

We can rewrite the equation as \begin{align}x^2+2x+3&\equiv0\mod198\\
(x+1)^2-1+3&\equiv0\mod 198\\
(x+1)^2&\equiv-2\mod198\\
(x+1)^2&\equiv196\mod198\end{align}
We let $y=x+1$ and we now need to solve \begin{align}y^2&\equiv-2\mod 198\\
&\equiv196\mod198\end{align}
We can see that $196=14^2$ and so we can say that $y\equiv14$ is a solution. Therefore, we can also say that $y\equiv-14\mod198\equiv184$ is also a solution
So, we have found two values for $x$ so far: $x\in\{13,183\}$
We can use WolframAlpha to find that there are two more solutions for $y$: $y\equiv58$ and $y\equiv-58\mod 198\equiv140$ however I'm not entirely sure how you compute these without using trial and error
This gives us the final two values for $x$: $x\in\{13,57,139,183\}$
