<s>[INST] 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
 [/INST] 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.
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