Conventions: $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.
Consider the algebraic groups $ \mathbb{G}_a$ and $ \mathbb{G}_m$. They are the only one-dimensional connected groups and they are both solvable. What about two-dimensional connected groups? It turns out that if $ \mathrm{dim} G\leq 2$ then $ G$ is solvable.
For $ \mathrm{dim} G= 3$, this is no longer true, for instance $ \mathrm{SL}_2$ is $ 3$-dimensional but not solvable since it is perfect, i.e. equal to its commutator subgroup. So let's prove our theorem:
Consider a maximal torus $ T\subseteq G$. There are two possibilities: $ \mathrm{dim} T = 0$ or $ \dim T = 1$. If $ \dim T = 0$ then $ B$ is unipotent, and the conjugates of $ B$ cover $ G$ so $ G$ is unipotent, hence nilpotent, hence solvable.
This leaves the case $ \dim T = 1$. Then $ B$ is equal to its semisimple part, which implies that $ G = B$, and hence $ G$ is solvable. .
In this proof, we used that for a connected group $ G$, if $ B$ is either semisimple or unipotent then actually $ G = B$. We will prove this in the next post.