My matrix is
\begin{pmatrix} 3 & 4 & 5 \\ -2 & 7 & 3 \\ 5 & -8 & -3 \end{pmatrix}
Through the rule of Sarrus, I know (approximately) $\lambda_1 = 5.9$ $\lambda_2 = 3.5$ and $\lambda_3 = -2.4 $.
If someone could find $v_1$ I believe I can go from here to get the remaining $2$.
On
The eigenvectors will come from solving the equation $$\left[ \begin{array}{ccc} 3 & 4 & 5 \\ -2 & 7 & 3 \\ 5 & -8 & -3 \end{array}\right]\mathbf{v} = \lambda \mathbf{v}$$ for $\mathbf{v}$ where $$\mathbf{v}=\left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right]$$ Simply compute the matrix multiplication of $$\left[ \begin{array}{ccc} 3 & 4 & 5 \\ -2 & 7 & 3 \\ 5 & -8 & -3 \end{array}\right]\left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right]$$ and set it equal to $$\left[ \begin{array}{c} \lambda v_1 \\ \lambda v_2 \\ \lambda v_3 \end{array}\right]$$ to get a set of three equations. This will give you enough information to solve for $v_1,v_2$ and $v_3$, which will give you the eigenvector $\mathbf{v}$ that corresponds to the eigenvalue $\lambda$.
Step 1. Find the characteristic polynomial $\phi(x)=\det(A-xI)=x^3-7x^2-2x+50.$
Step 2. Find the roots of $\phi(x).$ According to Wolfram Alpha, these are
and
These are the eigenvalues of the matrix. (note that despite the presence of $i$, these are all real numbers).
Step 3. For each eigenvalue $x$ from Step 2, solve the system of linear equations with matrix form $(A-xI)X=0$ (i.e. find the kernel (AKA nullspace) of $A-xI$). The solutions $X$ are the eigenvectors of $A$ for the eigenvalue $x.$
Step 4. Cry at the thought of even attempting Step 3.
Step 5. Check to see if you've got the matrix right. If no, start over with the correct matrix. If yes, get a computer to find eigenvalues and eigenvectors using software such as Matlab instead, or ask Wolfram Alpha to do that instead of asking it to find the characteristic polynomial. (I'm not aware of any simple methods for finding eigenvectors if you only know the approximate eigenvalues.)