I am trying to understand the following proof of solvability for the group $B_2$.
Let $B_2 = \{\begin{pmatrix}a&b\\&d\end{pmatrix}:ad\in\Re^x, b\in\Re\}$
Let $U_2 = \{\begin{pmatrix}1&b\\&1\end{pmatrix}:b\in\Re\}$
$U_2$ normal subgroup as it is the kernel of $f:B_2\rightarrow\Re^x\times\Re^x$ defined as follows:
$f(\begin{pmatrix}a&b\\&d\end{pmatrix})=(a,d)$
$Ker(f)$ solvable since $U_2$ is nilpotent/abelian
$Im(f)$ commutative since its isomorphic to $\Re^x\times\Re^x$
But why do these two facts about the kernel and image of $f$ imply that $B_2$ is solvable?