Q: non-linear congruence I dont understand any step of the solution in this question below. I know that they are allowed to bring 7 to the other side but anything after that I dont understand. It would help if greater detail on each step can be provided so it can make sense as to what is happening and why it is.

I am not understanding what is going on in the table they have made...Any idea what they are doing in that table?

A: Nothing fancy. $-7 + 10 = 3$, so $-7 \equiv 3 \pmod 5$.

A: Recall $\ a\equiv b\pmod{5}\iff 5\ |\ a-b,\ $ so $\ -7\equiv 3\pmod{5}\ $ by $\ 5\ |\ {-}7-3$.
Simpler than that above, to solve $\rm\:x(x+3)\equiv 3\pmod 5\:$ note that
$$\rm\ \ \ mod\ 5\!:\ \{(x,x+3)\} \equiv \{(0,3),(1,4),(2,0),(3,1),(4,2)\}$$
and only the final two pairs have product $\equiv 3\pmod 5$
