Solve the recurrence relation $a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$, $a_0=0$, $a_1=1$ My task:
$a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$,  $a_0=0$, $a_1=1$
My solution
$x^{2}-6x+9$
$\Delta=0$
$x_0=3 $
So I am gonna use following formula:
$a_n=ar^{n}+bnr^{n}$
$a_n=a*(3)^{n}+bn*3^{n}$
$-8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=6b_{n-1}-9b_{n-2}$
$$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+6c$$ I am setting $c=\frac{4}{3}$ so 
$$b_n=6b_{n-1}-9b_{n-2}\implies\exists a,\,b:\,b_n=a*3^{n}+bn*3^{n}.$$From $b_0=\frac{4}{3},\,b_1=\frac{7}{3}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.
$$a=\frac{4}{3}$$
$$b=-\frac{5}{9}$$
$b_2=22$
$a_2=22-\frac{4}{3}=\frac{62}{3}$
Actual $a_2=-2$
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
Can I use this $b_n$ method if delta equals 0? or should $c=-\frac{4}{3}$?

I think that your choice of $c$ is wrong 
From $$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+4c$$
this gives you $c=2$ and $b_n=a_n+2$
$$b_0=2, b_1=3$$  after solving $$b_n=a*3^{n}+bn*3^{n}$$ you get $a=2 ,b=-1$
