We will find the solution to the following lhcc recurrence:
$a_{n} = -6 a_{n-1} - 9 a_{n-2} \text{ for } n \geq 2$ with initial conditions $a_0 = 3 , a_1 = 5$. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format $a_n = r^n$. (We assume also $r \neq 0$). In this case we get: $$r^n = -6 r^{n-1} - 9 r^{n-2}.$$ Since we are assuming $r \neq 0$ we can divide by the smallest power of r, i.e., $r^{n-2}$ to get the characteristic equation: $$r^2 = -6 r - 9.$$ (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.)
This characteristic equation has a single root $r$. (We say the root has multiplicity 2). Find r.
$r=$

Since the root is repeated, the general theory (Theorem 2 in section 5.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: $$a_n = \alpha_1 (r)^n + \alpha_2 n(r)^n$$ for suitable constants $\alpha_1, \alpha_2$.
To find the values of these constants we have to use the initial conditions $a_0 = 3, a_1 = 5$. These yield by using n=0 and n=1 in the formula above: $$3 = \alpha_1 (r)^0 + \alpha_2 0(r)^0$$ and $$5 = \alpha_1 (r)^1 + \alpha_2 1(r)^1$$ By plugging in your previously found numerical value for $r$ and doing some algebra, find $\alpha_1, \alpha_2$:
$\alpha_1 =$
$\alpha_2 =$

Note the final solution of the recurrence is: $$a_n = \alpha_1 (r)^n + \alpha_2 n(r)^n$$ where the numbers $r, \alpha_i$ have been found by your work. This gives an explicit numerical formula in terms of n for the $a_n$.