Solving for eigenvalues:
\begin{gather*}
(4-r)(-6-r)+24 = 0\\
\implies - 24 - 4r + 6r + r^2 = 0\\
\implies r^2 + 2r = 0 \implies r(r+2) = 0
\end{gather*}
eigenvalues are $0$ and $-2$.
When $r = 0$, Nullspace of $\begin{pmatrix}
4 & -3 \\
8 & -6
\end{pmatrix}$ equals to Nullspace of $\begin{pmatrix}
4 & -3 \\
0 & 0
\end{pmatrix}$, which is equal to the span of $\begin{pmatrix}
3 \\
4
\end{pmatrix}$.
When $r = -2$, Nullspace of $\begin{pmatrix}
6 & -3 \\
8 & -4
\end{pmatrix}$, which is equal to Nullspace of $\begin{pmatrix}
6 & -3 \\
0 & 0
\end{pmatrix}$, which is a span of $\begin{pmatrix}
1 \\
2
\end{pmatrix}$.
Then
$$\mathbf{x} = c_1\begin{pmatrix}3\\4\end{pmatrix} + c_2e^{-2t}\begin{pmatrix}1\\2\end{pmatrix}$$
As $t$ goes to infinity, the solutions tend to $c_1\begin{pmatrix}3\\4\end{pmatrix}$. A sketch of the phase portrait:
