Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix. 
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4  \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.

How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1    \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1    \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.

The characteristic polynomial of $A$ is $x^2-6x-1$, whose roots are $3\pm\sqrt{10}$. An example of an eigenvector with norm $1$ whose eigenvalue is $3+\sqrt{10}$ is $\frac1{\sqrt{20-2\sqrt{10}}}\bigl(\sqrt{10}-1,3\bigr)$ and an example of an eigenvector with norm $1$ whose eigenvalue is $3-\sqrt{10}$ is $\frac1{\sqrt{20-2\sqrt{10}}}\bigl(3,1-\sqrt{10}\bigr)$. So, take$$P=\frac1{\sqrt{20-2\sqrt{10}}}\begin{pmatrix}\sqrt{10}-1&3\\3&1-\sqrt{10}\end{pmatrix}$$and then $P^TAD=\left(\begin{smallmatrix}3+\sqrt{10}&0\\0&3-\sqrt{10}\end{smallmatrix}\right)$.
