I have the following Matrix:
$$B= \begin{bmatrix} 8 & -6 & -6 \\ 30 & -22 & -30 \\ -30 & 30 & 38 \end{bmatrix} $$
I find the characteristic polynomial as:
$$det(B-\lambda E)=-(\lambda-8)^3$$
So we have the root 8 (with multiplicty 3). Therefore $B$ has the eigenvalue 8
Now I solve for
$$(B-\lambda E)x= \begin{bmatrix} 8-8 & -6 & -6 \\ 30 & -22-8 & -30 \\ -30 & 30 & 38-8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$
$$0x-6y-6z=0$$ $$30x-30y-30z=0$$ $$-30x+30y+30z=0$$
I find the solutions to the equations as:
$$x=0$$ $$y=-1$$ $$z=1$$
Am I going in the right direction? What does the results tell me about which eigenvectors exist for the matrix B?
What you've done so far is find an eigenvector corresponding to the eigenvalue $\lambda=8$. This doesn't tell you anything about other possible (linearly independent) eigenvectors. To determine their existence, you'll need to determine $\mathrm{dim}N(\lambda I - B)$, where $N(\cdot)$ denotes the null space. Looking at the linear system, we can see that we must have $x=0$. Then, we can easily see that, though there are infinitely many solutions, they are all of the form $$v=\begin{pmatrix} 0\\ v_2\\ -v_2 \end{pmatrix}.$$ Therefore, there is only the one linearly independent eigenvector (i.e., $\mathrm{dim}N(\lambda I - B) = 1$).