Are polynomials over $\mathbb{R}$ solvable by radicals?

Question

I know that if $f$ is a polynomial over a subfield $F$ of $\mathbb{C}$ and $f$ is solvable by radicals, then the Galois group of $f$ over $F$ is solvable. I've also seen many applications of this fact in demonstrating that certain quintic polynomials over $\mathbb{Q}$ are not solvable by radicals. My question is this: does it follow that there exists a polynomial over $\mathbb{R}$, or for that matter $\mathbb{C}$, that is not solvable by radicals? If this is the case, to prove it is it simply a matter of finding a polynomial over $\mathbb{R}$ whose Galois group is not solvable? Any help is appreciated. Thanks!

$\textbf{Edit:}$

Given Steven Stadnicki's comment, I'll include precisely what I mean by "solvable by radicals". Here is the definition to which I refer in my question:

Let $f$ be a polynomial over a subfield $K$ of $\mathbb{C}$, and let $\Sigma$ be the splitting field of $f$ over $K$. We say $f$ is solvable by radicals if there exists a subfield $M$ of $\mathbb{C}$ containing $\Sigma$ such that $M:K$ is a radical extension.

Answer 1

The Galois group of a polynomial over $\mathbb{R}$ has order at most $2$. If a polynomial does not already split over $\mathbb{R}$, we adjoin a complex root and obtain $\mathbb{C}$, which is algebraically closed and hence is the splitting field of the polynomial. Thus every polynomial over $\mathbb{R}$ is solvable by radicals.

Answer 2

Every polynomial with real coefficients splits over $\mathbb C = \mathbb R[\sqrt{-1}]$. In particular, every polynomial with real coefficients is solvable by radicals (over $\mathbb R$).