Let $p(x) = ax^2 + bx + c \in \mathbb{R}[x]$ be a degree 2 polynomial with real coefficients such that $D := b^2 - 4ac$. I'd like to examine the structure of the ring $\mathbb{R}[x]/(p(x))$ in the following three cases: $D > 0, D < 0,$ and $D = 0$. (Note that $(p(x))$ is the ideal generated by $p(x)$.)
Where I Am:
Well, if $D>0$ (I think I've got this case)...
...then $p(x)$ has two distinct (real) roots; call them $\alpha$ and $\beta$. Therefore, we can write $p(x) = (x - \alpha)(x - \beta)$. So, by the Chinese Remainder Theorem, we have that $$ \mathbb{R}[x]/(p(x)) \cong \mathbb{R}[x]/(x - \alpha) \times \mathbb{R}[x]/(x - \beta). $$ Now, we claim that $\mathbb{R}[x]/(x - \alpha) \cong \mathbb{R}$. Indeed, consider the homomorphism $\phi:\mathbb{R}[x] \to \mathbb{R}$ given by $\phi(q(x)) = q(\alpha)$. Then, since $$ \phi(q(x)) = q(\alpha) = 0 \iff q(x) \in (x - \alpha), $$ we see that $\ker(\phi) = (x - \alpha)$. Furthermore, since we may consider every real number to be a constant polynomial, we see that $\phi$ is surjective. Thus, it follows from the First Isomorphism Theorem that $$ \mathbb{R}[x]/\ker(\phi) = \mathbb{R}[x]/(x - \alpha) \cong \phi(\mathbb{R}[x]) = \mathbb{R}. $$ Similarly, $\mathbb{R}[x]/(x - \beta) \cong \mathbb{R}$. Therefore, $$ \mathbb{R}[x]/(p(x)) \cong \mathbb{R} \times \mathbb{R}. $$
Now, if $D < 0$...
...then $p(x)$ has no real roots. Thus, $p(x)$ is irreducible. So, $\mathbb{R}[x]/(p(x))$ is certainly a field of some sort. I know that $\mathbb{C} \cong \mathbb{R}[x]/(x^2+1)$; but is this true of any quadratic irreducible in $\mathbb{R}[x]$? I would think not... but I'm having trouble describing this ring any further than this. (Note that I don't know anything about "field extensions"...)
And, finally, if $D=0$...
...then $p(x)$ has a single (real) root of multiplicity two; call it $\gamma$. Therefore, we can write $p(x) = (x - \gamma)(x - \gamma)$. May I make the same conclusion here that I did from the first case? I don't see why not, but I just don't feel all that confident doing so...
For $D>0$, you're entirely right.
For $D<0$, you actually do get $\Bbb C$ no matter what. We see this if we use the isomorphism $\Bbb R[x]/(p(x))\to\Bbb C$ which maps $1$ to $1$ and $x$ to one of the complex roots of $p$ (you will, of course, need to show that this is well-defined and an isomorphism).
For $D=0$, you get what is called the dual numbers. Just like $\Bbb C$ has a standard representation as $\Bbb R[i]/(i^2+1)$, the dual numbers have a standard representation as $\Bbb R[\epsilon]/(\epsilon^2)$. Again, for any other $p$, you can construct an isomorphism. This time $\Bbb R[\epsilon]/(\epsilon^2)\to\Bbb R[x]/(p(x))$ by sending $1$ to $1$ and $\epsilon$ to one of the linear factors of $p$ (again, show that it is well-defined and an isomorphism).