What does it mean for a polynomial to be solvable by radicals?

It means that the answer can be computed using only the operations $+$, $-$, $\times$, $\div$, and $n$th roots, in a finite number of steps, using the coefficients of the equation.

Given a finite field extension $F \subseteq K$. A $\textbf{root tower}$ of $K$ over $F$ is a finite sequence of extensions

$$F = K_1 \subseteq K_2 \subseteq \dots \subseteq K_n = K$$

Such that for every $i$ where $1 \le i \le n - 1$, there exists a prime $p_i$ and an element $q_i \in K_{i+1}$ such that $q_i^{p_i} \in K_i$ and $q_i \notin K_i$.

If $f(x) \in F[x]$ where $char F = 0$, we say that the equation $f(x) = 0$ is solvable by radicals, if there exists a finite extension of the splitting field of $f(x)$ that has a root tower over $F$.