We will use the Lax-Milgram Theorem. A weak solution of your problem is a $u\in H^2(\Omega)$ such that
$\begin{eqnarray} \Delta^2u=f & \Rightarrow & [\Delta^2u]\varphi=f\varphi\\ & \Rightarrow & \int_{\Omega}\Delta(\Delta u)\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & -\int_{\Omega}\nabla(\Delta u)\nabla\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & \int_{\Omega}\Delta u\Delta\varphi=\int_{\Omega} f\varphi, \end{eqnarray}$ for all $\varphi\in H^2_0(\Omega)$. Define the bilinear operator $B:H^2_0(\Omega)\times H^2_0(\Omega)\rightarrow\mathbb{R}$, $B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi.$ Statement 1 This bilinear operator is continuos.
In fact,
$\begin{eqnarray} |B(u,\varphi)| & \leq & \int_{\Omega}|\Delta u||\Delta\varphi|\\ & \leq & \|\Delta u\|^2_{L^2(\Omega)}\|\Delta \varphi\|^2_{L^2(\Omega)}\\ & \leq & C\|u\|^2_{H^2_0(\Omega)}\|\varphi\|^2_{H^2_0(\Omega)} \end{eqnarray}$ You can prove easily this last inequality.
Statemant 2 The bilinear operator is coercive.
In fact, ([Edited]be cautious: this step is highly nontrivial as pointed out in the comment) $B(u,u)=\int|\Delta u|^2=\color{blue}{\|\Delta u\|^2_{L^2(\Omega)}\geq C\|u\|^2_{H^2_0(\Omega)}}.$
We used that $\|\Delta u\|_{L^2(\Omega)}$ defines a norm on $H^2_0(\Omega)$ equivalent to the usual norm.
Then, by the Lax-Milgram Theorem, for each $f\in H^2_0(\Omega)$, exists an unique function $u\in H^2_0(\Omega)$ such that $B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi=\int_{\Omega}f\varphi,$ for all $\varphi\in H^2_0(\Omega)$.
I hope I help you.