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 $x^2+2x+3 \equiv0\mod{12}$
 A: Substitute $y=x+1$. Now the equations is transformed to $y^2+2\equiv 0 \,mod\,198$. As $196=14^2$,we have  $x=y-1=13$ is a soution.
A: Here's a more-or-less generalizable, manual way of finding all of the solutions:
First, as P. Vanchinathan does, change variable to $a := x + 1$, which transforms the equation into one with zero linear term:
$$a^2 + 2 \equiv 0 \pmod {198} .$$
(This step is option, but reduces the amount of later work.)
Now, we exploit the prime factorization $198 = 2 \cdot 3^2 \cdot 11$. Reducing modulo $2$ gives $a^2 \equiv 0 \pmod {198}$, so any solution $a$ to the above display equation has the form $a = 2b$. Substituting into the previous display equation gives
$$4 b^2 + 2 \equiv 0 \pmod {198},$$
which is equivalent to $$2 b^2 + 1 \equiv 0 \pmod {99}.$$
Now reducing modulo $11$ (and multiplying by $6$ to produce a monic polynomial on the l.h.s.) leaves
$$b^2 + 6 \equiv 0 \pmod {11} .$$
The l.h.s. factors as $(b + 4)(b - 4)$, so any solution $b$ to the equation modulo $99$ has the form $$b = 11 c \pm 4 .$$
Substituting in the above equation modulo $99$ and proceeding as before (in particular, multiplying both sides of the resulting equation by $7$, which is coprime to $9$ and hence produces an equivalent equation) gives
$$c^2 \pm 4 c + 3 \equiv 0 \pmod 9 .$$
We may factor the left-hand side as $(c \pm 1)(c \pm 3)$. Since the prime factorization of $9$ is $9 = 3^2$, the equation in $c$ has a solution iff either factor is $0$ or both of the above factors are congruent to $0$ modulo $3$. The latter is impossible since the difference of those factors modulo $3$ is $\pm 1$, so the solutions are exactly
$c = \mp 1, \mp 3$. Substituting these four solutions successively into our equations for $b, a, x$ gives all of the solutions to the original equation: $$x \equiv 13, 57, 139, 183 ,$$
which agrees with the solution lioness99a produced using W.A.
A: 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 $x$: $x\in\{13,57,139,183\}$
