Is $\mathbb{R}[X]/(P)$ isomorphic to $\mathbb C$ for every irreducible polynomial $P$ of degree $2?$

You're making working with the quotient ring harder than it needs to be. You're familiar with working with the usual presentation of the complex numbers $\mathbb{R}[i] / (i^2 + 1)$, right? Working in $\mathbb{R}[x] / (P)$ isn't much different; it's just $x^2$ reduces to something else.

It might help you to define a new variable $\xi$ to be the image of $x$ in $\mathbb{R}[x] / (P)$, and do arithmetic in terms of $\xi$.

Anyways, it's much, much easier to figure out the image $x$ in $\mathbb{C}$ than it is to find a preimage of $i$ in $\mathbb{R}[x] / (P)$. (the latter shouldn't be all that hard once you get used to arithmetic, though)

Hint: for it to be a homomorphism, you need to find a complex number α such that $\alpha^2 + b \alpha + c = 0$.

Hint $\ $ For an explicit isomorphism proceed as in the quadratic formula. Suppose R is an ordered field which is closed under taking sqrts of positive elements, and suppose that some polynomial $\rm\ a\:X^2 + b\:X + c \in R[X]\: $ of negative discriminant $\rm\:d = b^2 - 4ac < 0\:$ has a root $\rm\:x\:$ in some extension field $\rm\:F\supset R\:.$ Then $\rm\:-1\:$ is a square in $\rm\:F\:$ since $$\rm (b+2ax)^2 =\: b^2+4a(bx+ax^2) =\: b^2+4a(-c) =\: d < 0\ \ \Rightarrow\ \ \left(\frac{b+2ax}{\sqrt{-d}}\right)^2 =\: -1 $$

If $\mathbb R[X]/(P)$ injects into $\mathbb C$ (and the injection preserves $\mathbb R$), then $X$ must map to one of the complex roots of $P$ which you can find using the quadratic formula.

So if $p+qi$ is a root of $P$, then the residue class $[\frac{1}{q}X-\frac{p}{q}]$ ought to behave like $i$. Proving that is probably just a matter of inserting the quadratic formula (which splits nicely into real and imaginary terms here) and reducing.