My question is straightforward to pose: given a polynomial $f$ over a subfield of $\mathbb{C}$, are there conditions which guarantee the existence of a closed formula for the roots of $f$ in terms of its coefficients? $\textit{Edit:}$ By "closed formula", I am referring to a formula using addition, subtraction, multiplication, division, and extraction of roots, such as the quadratic formula.
In case my background in mathematics is relevant to anyone who may wish to answer my question, I just finished my first undergraduate course in Galois theory, in which I was introduced to the following theorem:
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.
This theorem offers us a method to demonstrate, for example, that the general quintic polynomial is not solvable by radicals. However, I did not encounter any theorems that address the question I stated above. Any answers or comments pertaining to my question will be much appreciated. Thanks.
In characteristic $0$ the theorem you're speaking of is actually an if and only if:
Theorem: If $E$ is a splitting field of $f \in k[x]$, where $k$ is a field of characteristic $0$, then $f(x)$ is solvable by radicals if and only if the Galois group of the extension $E/k$ is a solvable group.
In positive characteristic it's not an if and only if. The direction $f(x)$ solvable $\Rightarrow$ Galois group solvable remains true but the converse can fail. The counterexample is to consider $f(x) = x^p - x - t$ as a polynomial in $\mathbb F_p(t)[x]$. It can be shown that the Galois group of the splitting field is $\mathbb Z/p$ but $f$ is not solvable by radicals.
For a reference: that example and proofs of the theorems mentioned are all in Rotman's Advanced Modern Algebra.
Edit: What solvable by radicals means is that there is a formula for the roots which uses field operations, radicals, and elements of the base field. If you choose your base field to be $\mathbb R$ then those operations give you $\mathbb C$ and of course you get all the roots of your polynomial. The key here is that you don't want to take all elements from $\mathbb R$, you want to just use coefficients from the polynomial $f$. So in the theorem above what you need is to choose your base field $k$ to be the subfield of $\mathbb R$ generated by the coefficients of $f$. Then the theorem gives you what you want.