I've thought about this problem a bit, but I may be heading in the wrong direction. I know that if $f$ is a polynomial of prime degree over $\mathbb{Q}$ with precisely two nonreal zeros, then the Galois group of $f$ is isomorphic to $\mathbb{S}_p$. It follows that this polynomial's Galois group is isomorphic to $\mathbb{S}_3$, from which we can can conclude that the Galois group is solvable. Thus $f$ is solvable by radicals, meaning there exists some field $M$ containing $\Sigma$, the splitting field of $f$, such that $M:\mathbb{Q}$ is a radical extension. From here, I don't know how I can draw any conclusions about $\Sigma$ itself. Any help is appreciated.
Radical Extension: An Extension $L:K$ in $\mathbb{C}$ is radical if $L=K(\alpha_1,...,\alpha_{m})$ where for each $j=1,...,m$, there exists an integer $n_j$ such that $\alpha_j^{n_j}\in K(\alpha_1,...,\alpha_{j-1})$ for $j\geq 2$.
It is not necessarily true that everything expressible by the radicals contained in $M$ will be contained in $\Sigma$. So if we have $m = \sqrt{6}+\sqrt[4]{1+\sqrt[3]{4}}$, then letting $\alpha=\sqrt{6}$, $\beta=\sqrt[3]{4}$, and $\gamma=\sqrt[4]{1+\beta}$, we may create the radical extension $\mathbb{Q}(\alpha,\beta,\gamma)$. We also have $\mathbb{Q}(m)\subset \mathbb{Q}(\alpha,\beta,\gamma)$, but $\mathbb{Q}(m)$ is not a radical extension (at least that's my understanding - I just started learning this material so it's quite possible that I'm wrong).
Edit: I appreciate the answer already given, but it uses a fair amount of machinery I have yet to develop (quadratic subfields, Kummer field extensions, etc). Is there a more elementary route to take in attacking this problem?