Field $F[x]/\langle p(x)\rangle$ contains both roots of $p(x)$ if degree $2$

What they mean is the following. $E = F[y]/\langle p(y) \rangle$ is a field. (I have changed the name of the indeterminate to avoid clashes.) The map $\varphi : F \to E$ that sends $a \mapsto a + \langle p(y) \rangle$ is an injective homomorphism of rings with unity. Identify $F$ with the subfield $\varphi(F)$ of $E$. Then $p(x) \in F[x] \subseteq E[x]$.

Now we know that $c = y + \langle p(y) \rangle \in E$ is a root of $p(x) \in E[x]$ in $E$. Since the polynomial has degree $2$, $E$ will contain another (possibly not distinct from $c$) root $d$ of $p(x)$. This is simply because if you assume wlog that $p(x)$ is monic, since $x - c$ divides $p(x)$, we have $p(x) = (x-c)(x-d)$ for some $d \in E$.

I'd say this is phrased well enough. There are a few subtleties, but nothing to worry about.

One is that $p$ is, at first, a polynomial over $F$ and you form the quotient field $K = F[x]/\langle p(x)\rangle$. After that, you consider $p$ as a polynomial over $K$ that happens to have coefficients in the subfield (that is isomorphic to) $F$ of $K$. This polynomial $p$ obviously has a root in $K$, namely the residue class $\bar x$ of $x$.

Another subtlety is that the statement now talks about both roots of $p$, seemingly assuming that the other root somehow already exists (instead of saying that $p$ has two roots in $K$). Then again, considering $p$ as a polynomial over the algebraic closure $\bar K$, those other roots surely exists in $\bar K$, and then the formulation makes sense again.

Anyway, it is important to realize that in $K$ there is a root $d$ of $p$ that is not $\bar x$ (*), that it makes sense to consider $F(d)$ as a subfield of $K$, and that $F(d)$ is actually equal to $K$ (and not just isomorphic).

(*) ignoring for the sake of simplicity that $\bar x$ could be a double root.