I have to find an example of solvable Lie algebra $L$ such that the Killing form of $L$ isn't null. If we take the Borel subalgebra of $\mathfrak{sl}(2)$, we have that the Killing form of $L$ is the matrix $$ K= \left( \begin{array}{ccc} 0 & 0 \\ 0 & 4 \\ \end{array} \right) $$
Could you give me another example?
Take $\mathfrak{g}$ to be the algebra of $N\times N$ upper triangular matrices. Since the Lie bracket of two upper triangular matrices is a strictly upper triangular matrix (one with noughts on the leading diagonal) and such matrices also form a Lie algebra, so therefore we see that the first derived algebra $\mathfrak{g}^\prime = [\mathfrak{g},\,\mathfrak{g}]$ is the nilpotent Lie algebra of $N\times N$ strictly upper triangular matrices. So $\mathfrak{g}$ is solvable. Moreover, its Killing form does not vanish, so it itself is not nilpotent, a fact which you can also work out from first principles.
Now work out the Killing form for the explicit example of $3\times 3$ where we have the Lie algebra basis:
$$\hat{U}=\left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}\right);\;\hat{V}=\left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&0\end{array}\right);\;\hat{W}=\left(\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}\right)$$
$$\hat{X}=\left(\begin{array}{ccc}0&1&0\\0&0&0\\0&0&0\end{array}\right);\;\hat{Y}=\left(\begin{array}{ccc}0&0&0\\0&0&1\\0&0&0\end{array}\right);\;\hat{Z}=\left(\begin{array}{ccc}0&0&1\\0&0&0\\0&0&0\end{array}\right)$$
The derived algebra $\mathfrak{g}^\prime = [\mathfrak{g},\,\mathfrak{g}]$ is spanned by $\hat{X},\,\hat{Y},\,\hat{Z}$ and is the Heisenberg algebra with brackets $[\hat{X},\,\hat{Y}]=\hat{Z};\;[\hat{Z},\,\hat{X}]=[\hat{Z},\,\hat{Y}]=0$. If you work out the matrix of the Killing form for the basis in the order $\hat{U},\,\hat{V},\,\hat{W},\,\hat{X},\,\hat{Y},\,\hat{Z}$ then it is:
$$K=\left( \begin{array}{cccccc} 2 & -1 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 & 0 \\ -1 & -1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$
I have a suite of Mathematica Lie algebraic tools, so best not to try this at home without something similar unless you want to get really bored. However, if you're new to all this, actually writing your own Killing form and adjoint representation calculator in Mathematica is definitely something worthwhile to do as a learning exercise.
As another non null Killing form for solvable but not nilpotent example, see this answer by Dietrich Burde. He gives a family of 3 dimensional Lie algebras that are all solvable but not nilpotent algebras. All of the family have non null Killing forms aside from one member, which is a solvable algebra that is not nilpotent yet has a null Killing form.
One way to go about generating lots of examples is as follows: observe that for a Lie algebra $L$ and $x \in L$, by definition of the Killing form $\langle \cdot,\cdot \rangle$ we have $$\langle x,x \rangle=\mathrm{trace}_L(\mathrm{ad}(x)^2)=\lambda_1^2+\dots+\lambda_n^2,$$ where $\lambda_1,\dots,\lambda_n$ are th eigenvalues of $\mathrm{ad}(x)$ on $L$ (with multiplicities). So any solvable Lie algebra that contains an element $x$ such that $\mathrm{ad}(x)$ is not nilpotent and has all eigenvalues real will give you an example of the type you seek. In particular, the Borel subalgebra of any semisimple Lie algebra will work, since the coroots have this property. The examples given by @WetSavannaAnimal (+1) are essentially of this type (with "semisimple" replaced by "reductive"---of course what I wrote above is not quite true in this extra generality).
