Polynomial modulus Can anyone explain why the two solutions to $n^2+7n-2 = 0$ modulo $43$ are $n=13$ and $n=23$ and how they are found?

The number $43$ is a prime, thus we can work like we always do, for example, with real numbers: all the same rules apply to the integers modulo $43$. Now, we seek to complete the square. All equalities are in $\Bbb Z/43\Bbb Z$, the integers modulo $43$:
$${n^2} + 7n - 2 = 0$$
$${n^2} + 2\frac 72n - 2 = 0$$
Note that $2$ has as an inverse the number $22$ since $2\times 22=44=1 \pmod {43}$. Thus $\dfrac 7 2=7\times 2^{-1}=7\times 22=25$. Then we get 
$$\begin{align}n^2+2\cdot 25 n-2+25^2-25^2&=0\\
(n+25)^2&=25^2+2\\(n+25)^2&=25\\n+25&=\begin{cases}5\\-5\end{cases}\\n&=\begin{cases}-20=23\\-30=13\end{cases}\end{align}$$
