<s>[INST] Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$

My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is the solution in $\mathbb{Z}_{55}$
General solution is $x=-2+k55$.
But the answer is wrong..
 [/INST] $\rm mod\ 5\!:\ 0 \equiv x^2+2x+2 \equiv x^2-3x+2 \equiv (x-1)(x-2)\:\Rightarrow\: x\equiv \color{#C00}1,\,\color{#0A0}2.$
$\rm mod\ 11\!:\ 7x\equiv 3\equiv 14\:\Rightarrow\:x\equiv 2.\ $ Solving these congruences yields


*

*$\rm\ \ x\equiv 2\ mod\ 11,\ x\equiv \color{#0A0}2\ mod\ 5\,\Rightarrow \: x\equiv 2\ mod\ 55.\ $   

*$\rm\ \ x\equiv 2\ mod\ 11,\ x\equiv\color{#C00} 1\ mod\ 5\!:\, 1 \equiv x\equiv 2+11n\equiv 2+n\Rightarrow n\equiv -1,\,$ thus
$\rm\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ x = 2 + 11(-1 + 5k)\equiv -9\equiv 46\ mod\ 55.$
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