\begin{vmatrix} 7/2 & 2\\ 2 & 3\end{vmatrix}
The characteristic equation is |A - λ I | = 0
$(\frac{7}{2}-λ)(3- λ)-4=0 $
$ λ_1= \frac{13-\sqrt{65}}{4} $ $λ_2 = \frac{13+\sqrt{65}}{4} $
so (A- $\frac{13-\sqrt{65}}{4}$I)x=0
$ 2x_1 $ + ($ \frac{\sqrt{65}-1}{4})x_2 $
$x_1$ = $ \frac{1-\sqrt{65}}{8}x_2 $
this is so far what I got,,, and I want to know is it correct? then I will proceed λ2 thanks.
When I solve $(A-\lambda I){\bf{v}}=0$ to get the eigenvectors for $\lambda_1=\tfrac{1}{4}(13-\sqrt{65})$, I get \begin{align} 0&=\begin{pmatrix} \frac{7}{2}-\tfrac{1}{4}(13-\sqrt{65})&2\\ 2&3-\tfrac{1}{4}(13-\sqrt{65}) \end{pmatrix}\begin{pmatrix} v_1\\ v_2 \end{pmatrix}\\ &=\begin{pmatrix} (\frac{7}{2}-\tfrac{1}{4}(13-\sqrt{65}))v_1+2v_2\\ 2v_1+(3-\frac{1}{4}(13-\sqrt{65})v_2 \end{pmatrix}, \end{align} and thus the two equations \begin{align} \frac{1}{4}(1+\sqrt{65})v_1+2v_2&=0\implies v_2=-\frac{1}{8}(1+\sqrt{65})v_1,\\ 2v_1+\frac{1}{4}(\sqrt{65}-1)&=0\implies v_1=-\frac{1}{8}(\sqrt{65}-1)v_2, \end{align} which gives me the eigenvector \begin{align} {\bf{v}}_1=\begin{pmatrix} 1\\ -\frac{1}{8}(1+\sqrt{65}), \end{pmatrix} \end{align} or equivalently \begin{align} {\bf{v}}_1=\begin{pmatrix} \frac{1}{8}(1-\sqrt{65})\\ 1 \end{pmatrix}. \end{align} Note that since the these two vectors lie each other's span, since multiplication by $\pm\frac{1}{8}(1\mp\sqrt{65})$ of one gives the other (I'm sure you can see this by just calculating). I don't know how you got that other equality, but I'm guessing you just miscalculated something somewhere. I hope this helps.