Why $\mathbb{C}$ is isomorphic to $\mathbb{R}[x]/<x^2+1>$
There is some vague intuition that $\mathbb{C}$ is involved from the fact that $x^2+1=0$ has no real solutions. This is a good starting point, but here we'll examine the above statement further. For the remainder of the post we won't be proving that $\mathbb{R}[x]/<x^2+1>$ is a field, I will only explain why it is isomorphic to the field $\mathbb{C}$. It is enough in this case to show what the addition and in particular the $\textit{multiplication}$ operations are like in both rings, to establish a map between them that preserves the operations, and show that this map is bijective. We can start by examining what $\mathbb{R}[x]/<x^2+1>$ looks like.
In this quotient ring, $x^2+1=0 \Rightarrow x^2=-1.$ Then if we take any polynomial in $\mathbb{R}[x]$ we can replace the $x^2$ terms with $-1$. For instance, the polynomial $x^3 = x^2*x = -x.$ Taking it further, we can see that there are only linear terms: if we have a polynomial of degree $n, n$ odd, given by $x^n+c = x^2*x^2*...*x^2*x + c = \pm x + c$. If $n$ is even, then $x^n + c = -1 + c.$ In either case it is linear. These terms can be written as $a + bx + <x^2+1>$
This is where things get especially interesting. I'll leave the addition operation out of this sketch; let's define the multiplication operation: with the linear terms from earlier, we multiply them to get $(a+bx)(c+dx) = ac+adx+cbx+bdx^2 = (ac-bd) + (bc+ad)x \mod(x^2+1).$ This looks a lot like the multiplication operation in the complex numbers.
This suggests a mapping $\phi: \mathbb{R}[x]/<x^2+1> \rightarrow \mathbb{C},\,\, \phi([f(x)]) = f(i)$. Under this map we have that $a+bx \mapsto a+bi,$ which is exactly what we expect from the previous paragraph. The kernel of this isomorphism is exactly whichever polynomial in $\mathbb{R}$ has $x^2+1$ as a factor ($\ker(\phi) = <x^2+1>$). The map is bijective since you can reach any complex number with this map, and only elements in the kernel are multiples of $x^2+1$ (so only the identity equivalence class gets sent to the identity of $\mathbb{C}$).
So we have that $\mathbb{C} \cong \mathbb{R}[x]/<x^2+1>.$