Question about recurrence relation problem. 
solve the following recurrence relation, subject to given initial conditions.
$a_{n+1} = 6a_n -9,$
$a_0 = 0,$
$a_1 = 3.$
Here is what I have done.
$a_{n+1} - 6a_n +9 = 0$
$a_n = r^n$
$r^{n+1} - 6r^n + 9 = 0$
$r + 6 = 0$
$r = -6$
$a_n$ = $X(-6)^n$
$a_1 = 3 = X(-6)^1$
$X = -1/2$
Therefore 
$a_n = -1/2(-6)^n$
Am I on the right track or are there any mistakes I made somewhere?
Edited: added the capture of the problem

$$\begin{align}
a_{n+1}&=6a_n-9\\
a_{n+1}-\frac95&=6a_n-\frac{54}5\\
&=6\left(a_n-\frac95\right)\\
u_{n+1}&=6u_n\\
u_n&=6u_{n-1}=6^2u_{n-2}=\cdots=6^{n-1}u_1\\
&=6^{n-1}\left(a_1-\frac95\right)\\
&=6^{n-1}\left(3-\frac95\right)\\
a_n-\frac95&=\frac65(6^{n-1})\\
&=\frac{6^n}5\\
a_n&=\frac15\left(9+6^n\right)\qquad\blacksquare
\end{align}$$
Check:
$$\begin{align}
6a_n-9&=\frac65(9+6^n)-9\\
&=\frac95+6(6^n)\\
&=\frac15(9+6^{n+1})\\
&=a_{n+1}
\end{align} $$
Hence formula is correct.
