Quadratic Congruence (with Chinese Remainder Thm) How do we solve quadratic congruences such as: 


  $x^2 \equiv11 \pmod{39}$


I know I must use the chinese remainder theorem with $p = 13, 3$ but I've only done linear examples and am unsure about how to do quadratic ones.
$$x^2 \equiv 11 \pmod{39} \implies x^2 \equiv 2 \pmod 3 \implies \text{No solution}$$

EDIT
In general, when you want to solve for $$x^2 \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^2 \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
For instance, if you have $x^2 \equiv 23 \pmod {77}$, then we need to look at $x^2 \equiv 23 \pmod 7$ and $x^2 \equiv 23 \pmod{11}$ i.e. $x^2 \equiv 2 \pmod 7$ and $x^2 \equiv 1 \pmod{11}$.
$$x^2 \equiv 2 \pmod7 \implies x \equiv \pm 3 \pmod 7$$ Similarly, $$x^2 \equiv 1 \pmod{11} \implies x \equiv \pm 1 \pmod{11}$$
Hence,
$$x \equiv 3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 45 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 67 \pmod{77}$$
$$x \equiv 3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 10 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 32 \pmod{77}$$
Hence,
$$x \equiv \pm 10, \pm 32 \pmod{77}$$
$x^2\equiv11(\pmod{39}$

$39$ is a composite number $3\cdot13$

so now problem is :

$x^2\equiv11\pmod3$ and $x^2\equiv11\pmod{13}$.

if $x^2\equiv a\pmod{p}$ is a problem then it has solution if $a^{((p-1)/2)} \pmod {p}=1$

and have no solution if ^((p-1)/2) mod(p)=p-1
1)x^2=11mod(3) (p-1)/2=1
 11^1 mod(3)=2=p-1 so no solution
2)x^2=11 mod(13) (p-1)/2=6
 11^6 mod(13)=12=p-1 so no solution