Borel subalgebras contain solvable radical

Take the commutator of this stuff with itself again and its an induction argument on the length of the derived series (the longer one of the two).

I just want to make this step in Ravi's answer clearer, since I don't think it is obvious (well, at least to me...).

Using the fact that $\text{Rad }L$ is an ideal, we have $$[B+\text{Rad }L,B+\text{Rad }L]\subset [B,B]+\text{Rad }L.\hspace{1cm}(*)$$ If we use the notation $L^{(0)}=L$, $L^{(n)}=[L^{(n-1)},L^{(n-1)}]$, then $(*)$ implies that for each $n$, $$[B^{(n-1)}+\text{Rad }L,B^{(n-1)}+\text{Rad }L]\subset B^{(n)}+\text{Rad }L.$$ So we can prove by induction that $$\begin{aligned}(B+\text{Rad }L)^{(n)}&\subset B^{(n)}+\text{Rad }L.\end{aligned}$$ Now, since $B$ is solvable, $B^{(n)}=0$ for some $n$, so $(B+\text{Rad }L)^{(n)}\subset \text{Rad }L$. Since $\text{Rad }L$ is solvable, we conclude that $(B+\text{Rad }L)^{(n)}$ is solvable. Thus $B+\text{Rad }L$ is solvable.

We have $$[B+\text {Rad} L, B+\text {Rad} L] = [B,B]+[B,\text{Rad} L]+[\text{Rad} L,B]+[\text{Rad} L,\text{Rad}L].$$

The first term in in the commutator of B, the second two terms are in Rad L and the last one is in the commutator of Rad $L$. Take the commutator of this stuff with itself again and its an induction argument on the length of the derived series (the longer one of the two).