Cardano's formulas gives us the roots of a cubic in terms of radicals. For a depressed cubic $x^3+px+q$, these roots are
$$\begin{equation} \label{equation:cubic-roots} r_1 = {z_1+z_2\over 3},\quad r_2 = {\omega^2 z_1 + \omega z_2 \over 3},\quad r_3 = {\omega z_1 + \omega^2 z_2 \over 3}, \end{equation}$$
where $\omega$ is a primitive 3rd root of unity, $d=-4p^3-27q^2$ is the discriminant and
$$z_1 = \sqrt[3]{-27q+3\sqrt{-3d} \over 2},\quad z_2 = \sqrt[3]{-27q-3\sqrt{-3d} \over 2}.$$
As \eqref{equation:cubic-roots} shows, the solution takes a detour to the complex numbers, even when all the roots are real. I'll show that when the cubic is irreducible (casus irreducibilis) and has real roots, this phenomenon is unavoidable (see the below Theorem). Thus one cannot hope to find some alternative expression for the roots that avoids complex numbers.
Let $F$ be a subfield of $\R$ and $f(x)\in F[x]$ be an irreducible cubic with real roots. The degree of the splitting field $E\subset\R$ is at least 3, so $G_f=A_3$ or $S_3$. Note also that if $d$ is the discriminant, then $\sqrt d\in E$ is real, and so $d>0$, which implies $G_f=A_3$.
Lemma. If $r$ is real and $r^p\in F$ for $p$ a prime, then $F(r)$ does not contain a splitting field of $f(x)$ over $F$. Furthermore, $f(x)$ remains irreducible over $F(r)$.
Proof. I proved in [this other entry]({% post_url 2023-01-25-galoisgrp-xp-a %}) that $x^p-r^p$ is either irreducible or has a root in $F$. Thus $[F(r):F]=1$ or $p$. Supposing that $F(r)$ contains a splitting field $E$, we can eliminate the first possibility and by degree considerations also conclude that $p=3$, so that $F(r)=E$. But this means $F(r)/F$ is normal, so it contains the conjugates $\omega r$, $\omega^2r$, where $\omega$ is a primitive 3rd root of unity. So $F(r)\not\subset\R$, implying $r$ is not real. Lastly, $f(x)$ must be irreducible over $F(r)$, otherwise $F(r)$ contains $F(r_1)$ for some root $r_1$, which is a splitting field since $[F(r_1):F]=3$. $\qed$
Theorem. There exists no subfield $K$ of $\R$ that has a root tower of $F$ and contains a splitting field of $f(x)$ over $F$.
Proof. Fix a root tower and suppose $L\subset L(\sqrt[n]{a})$ are successive elements, where $a\in L$. We can factor the extension $L(\sqrt[n]{a})/L$ into a tower of subextensions with prime indices. For example if $n=pq$ with $p,q$ prime then we have the tower
$$L\subset L(\sqrt[p]{a}) \subset L(\sqrt[p]{a},\sqrt[pq]{a}) = L(\sqrt[n]{a}).$$
Since $K\subset\R$, all the radicals are real. So if $L$ does not contain a splitting field of $f(x)$ and $f(x)$ is irreducible over $L$, then we can repeatedly apply the lemma to conclude that $L(\sqrt[n]{a})$ does not contain a splitting field either and $f(x)$ remains irreducible over $L(\sqrt[n]{a})$. By working our way up the root tower starting from $F$, we can conclude that $K$ does not contain a splitting field. $\qed$