If you want to solve the quintic over the $p$-adics, two cases naturally arise : $p\neq5$ and $p=5$.

Suppose first that $p\neq5$ and let $f\in\mathbf{Q}_p[T]$ be an irreducible polynomial of degree $5$. Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_p$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_p(\root5\of1)$. So you see immediately that $\alpha$ can be expressed by radicals.

In fact, $f$ can always be taken to be of the form $f=T^5-x$ in the generic case when $K$ is (totally) ramified over $\mathbf{Q}_p$, so if you wish I can claim to have solved the quintic by radicals by just saying that $\alpha=\root5\of x$.

Now let $f\in\mathbf{Q}_5[T]$ be an irreducible polynomial of degree $5$. Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_5$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_5(\root4\of{\mathbf{Q}_5^\times})$. Here again you see that $\alpha$ can be expressed by radicals.

There is nothing special about the prime $5$ or the base field $\mathbf{Q}_p$. You can replace $5$ by any prime $l$, and $\mathbf{Q}_p$ by any finite extension thereof, and you will get similar results depending on whether $l\neq p$ or $l=p$.

You can even replace $\mathbf{Q}_p$ by a finite extension of $\mathbf{F}_p((\pi))$, provided you replace the "radical" $\root p\of{x}$ (which denotes a root of the binomial $T^p-x$) by its characteristic-$p$ cousin $\wp^{-1}(x)$ (which denotes a root of the trinomial $T^p-T-x$).

doesn'tmean there is a formula for the zeros in terms of the coefficients of the original polynomial. $\endgroup$2more comments