In congruence classes Z/(mZ), reduce the equation a_m*x_m^2=c_m either by finding convenient representation for a_m and b_m or by using the inverse of

Step 1
Let ${\displaystyle y=x^2}$ then the given congruence relation becomes,
${\displaystyle ay\equiv b\text{ mod }\text{mt }\hat{i}s2y\equiv 13\text{mod}19}$
This congruence has unique solution if and only if gcd(a,m)=1
Here gcd(2,19)=1 hence the given congruence has unique solution.
Now since 2 has an inverse, we get ${\displaystyle y\equiv 2^{-1}13\text{mod}19}$ which is the only solution.
The inverse of ${\displaystyle 2\in {\mathbb{Z}}_{19}}$ is 10.
${\displaystyle y\equiv 130\text{mod}19\Rightarrow y=16}$
Now ${\displaystyle y=x^2\Rightarrow 16=4^2}$. Hence x =4 is the required solution.