Discrete math: $a_n=14a_{n−1}−33a_{n−2}$ for $n≥3$ with initial conditions $a_0=−24,a_1=−200$ Can someone please help me understand this? This is what I got but it isn't correct.
$a_n=14a_{n−1}−33a_{n−2}$ for $n≥3$ with initial conditions $a_0=−24,a_1=−200$.
Solve for $a_n$.
$t^2 = -14t+33$
$r_1=-11$
$r_2= -3$
$n_0 = M(-11)^0 + N(-3)^0 = -24$
$n_1 = M(-11)^1 + N(-3)^1 = -200$
By algebra, $m = 34 , n = -58$, so
$a_n= 34(-11)^n - 58(-3)^n$.

Just a sign error in the setup.  You suppose that $a_n = t^n$ is the general rule, and plugging into the recurrence relation (for $n \geq 3$) gives 
$$t^n = 14 t^{n-1} - 33 t^{n-2},$$
or, putting everything on one side,
\begin{align}
t^n - 14 t^{n-1} + 33 t^{n-2} &= 0\\
t^{n-2}(t^2 - 14 t + 33) & = 0\\
\end{align}
[I trust the nature of the sign error is already clear.]
Thus, for $n \geq 3$, we have that 
\begin{align}
t^{n-2} & = 0 && \text{or} && t^2 - 14t + 33 && = 0\\
t & = 0 && \text{or} && (t-11)(t-3) && = 0\\
t & = 0 && \text{or} && t -11 && = 0 && \text{or} && t-3 && = 0\\
t & = 0 && \text{or} && t && = 11 && \text{or} && t && = 3\\
\end{align}
and we see that the roots are $0$, $\mathbf{+}11$, and $\mathbf{+}3$.  I presume that $t = 0$ ends up having no role (since $0^n$ is just $0$ and it is just extra hassle to add it when it does not help).
