Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question:

Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$

My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$
I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this).
thanks in advance
 A: Let's take one equation
\begin{align*}
3x^2+7x+7 & \equiv 0 \pmod{11}\\
4(3x^2+7x+7) & \equiv 4(0) \pmod{11}\\
x^2+28x+28 & \equiv 0 \pmod{11}&& (\because 4(3) \equiv 1 \pmod{11})\\
x^2+6x+6 & \equiv 0 \pmod{11}&& (\because 28 \equiv 6 \pmod{11})\\
(x+3)^2-3 & \equiv 0 \pmod{11}\\
(x+3)^2-5^2 & \equiv 0 \pmod{11}&& (\because 5^2 \equiv 3 \pmod{11})\\
(x-2)(x+8) & \equiv 0 \pmod{11}
\end{align*}
Since $11$ is prime so if $11 | ab$, then $11$ divides at least one of them, so we get
$$x\equiv 2 \pmod{11} \quad \text{ or } \quad x \equiv -8 \equiv 3\pmod{11}.$$
Likewise (you can work this out yourself)
$$3x^2+5x+4 \equiv 0 \pmod{13} \implies x\equiv 2 \pmod{13} \, \text{ or } \, x \equiv \color{blue}{b}\pmod{13}. $$
So we have the following situation
\begin{align*}
x&\equiv 2 \pmod{11} & x&\equiv 2 \pmod{11} & x&\equiv 3 \pmod{11} & x&\equiv 3 \pmod{11}\\
x&\equiv 2 \pmod{13} & x&\equiv  \color{blue}{b} \pmod{13} & x&\equiv  \color{blue}{b} \pmod{13} & x&\equiv 2 \pmod{13}
\end{align*}
Now use CRT (hopefully you know how to apply it to simple systems like these) to solve these systems.
For example the last system
\begin{align*}
x & \equiv 3 \pmod{11}\\
x & \equiv 2 \pmod{13}
\end{align*}
yields
$$x \equiv 3(13)(6)+2(11)(6) \equiv \color{red}{80} \pmod{143}. $$
Likewise  you will get a total of $\color{red}{4}$ incongruent solutions.
A: hint
as
$$95\equiv -48 \mod 143$$
the equation becomes
$$3x^2+18x-48\equiv 0 \mod 143$$
but
$$\delta=81+144=225=(15)^2$$
thus, it gives
$$3(x-2)(x+8)\equiv 0\mod 143$$
A: HINT.-One has $3(x+3)^2+68\equiv0\pmod{11*13}$ so $$Y^2\equiv3\pmod{11}\\Y^2\equiv-1\pmod{13}$$ It follows $$Y\equiv5,6\pmod{11}\Rightarrow x\equiv2,3\pmod{11}\\Y\equiv5,8\pmod{13}\Rightarrow x\equiv2,5\pmod{13}$$
Thus $$x=2,57,80,135\pmod{143}$$ i.e. $x=2+143n$ for $n\in\mathbb Z$ and so is for $57,80$ and $135$.
A: You are allowed to use the quadraditic formula.
$3x^2 + 18x + 95 \equiv 0 \pmod {143}$ means
$x \equiv \frac {18\pm \sqrt{18^2 -4*95*3}}{6} \pmod {143}$.
Now use Chinese remainder theorem on $143 = 13*11$
$\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {11}\equiv$
$\frac {7\pm {7^2 - 4*7*3}}{6}*12\pmod {11}\equiv$
$2(7\pm \sqrt{49 -12*7}) \equiv 14\pm 2\sqrt {5-7}\equiv$
$3 \pm 2\sqrt {-2}\equiv 3\pm 2\sqrt{9,64} \equiv 3\pm 6,16\equiv 9,8\pmod{11}$
And $\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {13}\equiv$
$\frac {5\pm \sqrt {5^2 - 12*4}}{6}\pmod {13}\equiv$
$-12(\frac {5\pm \sqrt {25 + 4}}{6}\equiv -2(5\pm \sqrt{29})\equiv$
$-10 \pm2\sqrt{3}\equiv 3 \pm 2\sqrt{16,81}\equiv 3\pm 8,18\equiv 11,8$
So you have four solutions.
A: You can also perform all calculations in the $\text{modulo } 143$ system, but there is some algebra you have to use to verify you get all the solutions.
Using elementary number theory we find that $3^{-1} \equiv 48 \mod 143$, so multiplying through we can also, equivalently, examine
$\tag 1 x^2 + 6x + 127 \equiv 0 \mod 143$
and this can be written as
$\tag 2 (x + 3)^2 \equiv 25 \mod 143$
and we get an easy solution, $x + 3 \equiv 5 \mod 143$.
Using brute force logic we can write as true
$\quad x^2 \equiv 1 \mod 143 \iff$ 
$\quad\quad x \equiv 1  \mod 143 \,\lor\, x \equiv 12  \mod 143 \,\lor\, x \equiv 131  \mod 143 \,\lor\, x \equiv 142 \mod 143$
Multiplying out we get $4$ solutions to $u^2 \equiv 25 \mod 143$,
$\quad 5 \times 1 \equiv 5 \mod 143$
$\quad 5 \times 12 \equiv 60 \mod 143$
$\quad 5 \times 131 \equiv 83 \mod 143$
$\quad 5 \times 142 \equiv 138 \mod 143$
Thus,
$x=2,57,80,135 \mod 143$
