The notion of an equivalence class always arises when dealing with something called an "equivalence relation." So to explain the former, we'll first discuss the latter.
An equivalence relation is (speaking broadly) a generalized notion of equality. For example, if we are conducting a survey, and we are only concerned with gender, we might say that two people are "equivalent" if they have the same gender. Similarly, in the case of modular arithmetic, we say that two numbers are equivalent if they have the same remainder when divided by some number $n$. So, if we are working mod $4$, we say that $1 \equiv 5$. That does not mean that $1=5$, just that they are equivalent for our purposes.
More formally, an equivalence relationship needs to be reflexive (everything should be equivalent to itself), symmetric (if $A$ is equivalent to $B$, $B$ is equivalent to $A$) and transitive (if $A$ is equivalent to $B$, and $B$ is equivalent to $C$, $A$ is equivalent to $C$). Thus, this generalized notion of equality behaves very similarly to the traditional "$=$".
When we construct an equivalence relationship, one can think of it as putting on a pair of coarse glasses, so that we can no longer distinguish between "equivalent" objects. Thus, mod $4$, it appears to us that $1,5,9,\cdots$ are all the same thing. This entire family of objects that are equivalent is called an equivalence class. When we are working mod $4$, there will be $4$ equivalence classes, one corresponding to all things with remainder $0$, another for all things with remainder $1$, etc. Moreover, when you add two things with remainder $2$, you get something with remainder $0$. Thus, we can say that when working mod $4$, there are only four objects: families of "equivalent" numbers, and we do arithmetic with these families.
It seems to be the list of all polynomials of degree less than $2$, with coefficients in $\mathbf Z/3\mathbf Z$ in the first case; of all polynomials of degree less than $3$, with coefficients in $\mathbf Z/2\mathbf Z$ in the second case.
Indeed, for the first case, a polynomial of degree less than $2$ has the form $a_0+a_1x$. Plug in the different possible values for the pairs $(a_0,a_1)$, getting
\begin{matrix}
(0,0)\to \color{red}{0}&(0,1)\to \color{red}{x}&(0,2)\to \color{red}{2x}\\
(1,0)\to \color{red}{1}&(1,1)\to \color{red}{1+x}&(1,2)\to \color{red}{1+2x}\\
(2,0)\to \color{red}{2}&(2,1)\to \color{red}{2+x}&(2,2)\to \color{red}{2+2x}
\end{matrix}
The second case goe along the same lines:
\begin{matrix}
&(0,0,1)&&(0,1,1)\\
&\color{red}{x^2} &&\color{red}{x+x^2}\\
(0,0,0)&&(0,1,0) \\
\color{red}{0}&&\color{red}{x}\\[2ex]
&(1,0,1)&&(1,1,1)\\
& \color{red}{1+x}&&\color{red}{1+x+x^2}\\
(1,0,0)&&(1,1,0)\\
\color{red}{1}&& \color{red}{1+x}
\end{matrix}
Best Answer
We have $2x^2+8x+2\equiv 0\pmod{23}$ if and only if $x^2+4x+1\equiv 0\pmod{23}$.
Now complete the square. We have $x^2+4x+1=(x+2)^2-3$. So we want to solve the congruence $(x+2)^2\equiv 3\pmod{23}$.
Let $y=x+2$. We want to solve the congruence $y^2\equiv 3\pmod{23}$.
There is general theory that, for large $p$, helps us determine whether a congruence $y^2\equiv a \pmod{p}$ has a solution, and even to compute a solution. But at this stage you are probably expected to solve such things by inspection.
Note that $y\equiv 7\pmod{23}$ works, and therefore $y\equiv -7\equiv 16\pmod{23}$ also works. We have found two solutions, and by general theory if $p\gt 2$ there are either $2$ solutions or none, so we have found all the solutions.
From $y\equiv 7\pmod{3}$ we conclude that $x+2\equiv 7\pmod{23}$, and therefore $x\equiv 5\pmod{23}$.
From $y\equiv 16\pmod{23}$ we conclude that $x\equiv 14\pmod{23}$.
Remarks: If our congruence had been (for example) $x^2+7x-8\equiv 0\pmod{23}$, there would be a bit of unpleasantness in completing the square, since $7$ is odd. But we could replace $7$ by, say, $30$, and complete the square to get $(x+15)^2-225-8$. So our congruence would become $(x+15)^2\equiv 233\pmod {23}$, or equivalently $(x+15)^2\equiv 3\pmod {23}$.
In general, if we have $ax^2+bx+c\equiv 0\pmod{p}$, it is useful to multiply through by the inverse of $a$ modulo $p$ to make the lead coefficient equal to $1$. There are a number of other helpful "tricks."