When does a formula for the roots of a polynomial exist?

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.