System of (non-linear) congruence equations I got a system of two congruence equations where one of them is non-linear.
\begin{cases} 2*x^2 + 5 \equiv 4\ (\textrm{mod}\ 11) \\ x \equiv 3\ (\textrm{mod}\ 13) \end{cases}
My idea was to rewrite the first equation like this:
$2*x^2 \equiv 10\ (\textrm{mod})\ 11$
And then use the chinese reminder theorem to find solutions to this equations:
\begin{cases} x \equiv 10\ (\textrm{mod}\ 11) \\ x \equiv 3\ (\textrm{mod}\ 13) \end{cases}
This gives me the solution $x=120$ which does not work for the first equation of my original equations.
I then tried to add $11*13$ to this solution until the double of a square number occurs. But it looks like this doesn't work.
Can you give me some advice on how to solve such an equation?
 A: From $x\equiv 3\mod 13$ we conclude that $$x^2\equiv 9\mod 13$$define $y=x^2$, therefore $$2y+5\equiv 4\mod 11\\y\equiv 9\mod 13$$applying Chinese Remainder Theorem on the above set of equations we conclude that $$y\equiv126\mod 143$$therefore $$x^2\equiv 126\mod 143$$among which the answers with $x\equiv 3\mod 13$ are acceptable. Now let $$x=143k+r\quad,\quad 0\le r<143$$by substitution we obtain $$x^2\equiv r^2\equiv 126 \mod 143$$Numerical simulations show that the only possible values for $r$ are $$29\quad 62\quad  81\quad 114$$also the constraint $x\equiv 3\mod 13$ imposes that $$r\equiv 3 \mod 13$$ which leaves us with $$r=29\\r=81$$therefore we if an answer $x$ exists, it must be either$$x\equiv 29\mod 143$$or $$x\equiv 81\mod 143$$There is one final step. To show that these answers are also sufficient condition, we show that they satisfy the primary equations of the question. Let $x=143k+29$ therefore $$2x^2+1\equiv 2\times29^2+1\equiv 2\times (-4)^2+1\equiv 33\equiv 0\mod 11$$also for $x=143k+81$ we have $$2x^2+1\equiv 2\times81^2+1\equiv 2\times (-7)^2+1\equiv 99\equiv 0\mod 11$$

Conclusion
The only answers are$$x\equiv 29\text{ or }81\mod 143$$

A: HINT:
$$x\equiv3\pmod{13}\implies x-3=13l\\2x^2\equiv2(3+13l)^2\equiv18+156l+338l^2\equiv-1\pmod{11}\implies 8+2l-3l^2\equiv0\pmod{11}$$
