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 justify the statement. 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. 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 $x^n + c = \pm 1 + c.$ In either case it is linear. These terms can be written as $a + bx + <x^2+1>$
This is where things get especially interesting. Let's define the multiplication operation: with the linear terms from earlier, we multiply them to get $(a+bx + <x^2+1>)(c+dx + <x^2+1>) = ... = (ac-bd) + (bc+ad)x + <x^2+1>.$ Look familiar? This looks a lot like the multiplication operation in the complex numbers.
It turns out that the isomorphism between this ring and $\mathbb{C}$ is the mapping $f(x) + <x^2+1> \rightarrow f(i)$. Try replacing the above defined multiplication operation with $i$ wherever you have $x$ and observe that it is identical to the multiplication operation in $\mathbb{C}$. The kernel of this isomorphism is exactly whichever polynomial in $\mathbb{R}$ has $x^2+1$ as a factor. This is a rough sketch of the reason why. Further invesigation is needed on your part buddy.