How to solve second order congruence equation if modulo is not a prime number the equation is $x^2 = 57 \pmod{64}$
I know how to solve equations like 
(*) $ax^2 +bx +c = 0 \pmod{p}$, where $p$ is prime 
and i know all the definitions for like Legendre's Symbol and all of the other quadratic residue terminology and that (*) can be rewritten in form $x^2 = c \pmod{p}$ but here $64$ is $2^6$. I know that if $a = b \pmod{n}$ and $d|n$ then $a = b \pmod{d}$ but i'm not sure if i can apply this here 
can someone solve it without explaining related terminology like what is quadratic residue, excetera..
Thanks!

First, also from the reciprocity law, we know that $\;x^2=z\pmod{2^k}\;$ , has a solution for $\;k\ge 3\;$ iff $\;z=1\pmod 8\;$, which is the case here.
Now:
$$x_k^2=z\pmod{2^k}\implies x_{k+1}=\begin{cases}x_k+2^{k-1}&,\;\;\frac{x_k^2-z}{2^k}=1\pmod 2\\{}\\x_k&,\;\;\frac{x_k-z}{2^k}=0\pmod 2\end{cases}$$
and you can check easily $\;x_{k+1}\;$ is a solution to the equation modulo $\;2^{k+1}\;$
In our case:
$$x_3=1\;,\;\;\text{because}\;\;1^2=57\pmod 8\;,\;\;\text{and since}\;\;\frac{1^1-57}8=1\pmod 2$$
we get that
$$x_4=1+2^2=5\;\;\text{fulfills}\;\;5^2=57\pmod{16},\;\;\text{and since}\;\;\frac{5^2-57}{16}=0\pmod{16}$$
then also
$$x_5=5\;\;\text{fulfills}\;\;5^2=57\pmod{32},\;\;\text{and since}\;\;\frac{5^2-57}{32}=1\pmod 2$$
we get that
$$x_6=5+2^4=21\;\;\;\text{fulfills}\;\;\;21^2=57\pmod{64}$$
