Finding the closed form for a sequence My teacher isn't great with explaining his work and the book we have doesn't cover anything like this. He wants us to find a closed form for the sequence defined by:
$P_{0} = 0$
$P_{1} = 1$
$\vdots$
$P_{n} = -2P_{n-1} + 15P_{n-2}$
I'm not asking for a straight up solution, I just have no idea where to start with it. The notes he gave us say:

We will consider a linear difference equation that gives the
  Fibonacci sequence.
$y(k) + A_1y(k -1) + A_2y(k -2) = \beta$ 
That's the general form for a difference equation in which each term is formed from the
  two preceding terms. We specialize this for our Fibonacci sequence by setting $A_1 = 1, $ >$
>A_2 = 1,$ and $ \beta = 0$. With some rearrangement, we get
$y(k) = y(k - 1) + y(k - 2)$
which looks more like the general form for the Fibonacci sequence.
To solve the difference equation, we try the solution $y(k) = Br^k$. Plugging that in, we
  obtain
$Br^{k-2} (r^2 - r - 1) = 0$

I have no idea where the $Br^k$ is coming from nor what it means, and he won't explain it in any sort of terms we can understand.
If someone could help me with the basic principle behind finding a closed form with the given information I would be eternally grateful.

EDIT: Using the information given (thank you guys so much) I came up with 
$y(k) = \frac{1}{8}(3)^k - \frac{1}{8}(-5)^k$
If anyone has ran through let me know what you found, but I'm in no way asking you guys to do that. It's a lot of work to help some random college student.

A related problem. Here is a start. Just assume your solution $P_n=r^n$ and plug in back in the eq. to find $r$
$$ P_{n} = -2P_{n-1} + 15P_{n-2} \implies r^n+2r^{n-1}-15r^{n-2}=0  $$
$$ \implies r^{n-2}(r^2+2r-15)=0 \implies r^2+2r-15=0 $$ 
Find the roots of the above polynomial $r_1, r_2$ and construct the general solution
$$ P(n)=c_1 r_1^n + c_2 r_2^n \longrightarrow (*)  $$
To find $c_1$ and $c_2$, just use $P_0=0$ and $P_1=1$ in $(*)$ to get two equations in $c_1$ and $c_2$. Once you find $c_1$ and $c_2$ plug them back in $(*)$ and this will be the required solution. 
