A solvable version of the Baer-Suzuki theorem

Suppose that $G$ is a finite group and $x \in G$ has prime order $p \ge 5$. Then $x$ is contained in the solvable radical of $G$, $O_{\infty}(G)$, if (and only if) $\langle x,x^g \rangle$ is solvable for all $g \in G$. If $G$ is an almost simple group and $x \in G$ has prime order $p \ge 5$, then this implies that there exists $g \in G$ such that $\langle x,x^g \rangle$ is not solvable. In fact, this is also true when $p=3$ with very few exceptions, which are described explicitly.