How to solve a recursive equation I have been given a task to solve the following recursive equation
\begin{align*}
a_1&=-2\\
a_2&= 12\\
a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2, \quad n \geq 3.
\end{align*}
Should I start by rewriting $a_n$ or is there some kind of approach to solve these?
I tried rewriting it to a Quadratic Equation (English isn't my native language, sorry if this is incorrect). Is this the right approach, if so how do I continue?
\begin{align*}
a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2\\
x^2&= -4x-4\\
0&= x^2 + 4x + 4
\end{align*}

Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$. Rewrite your recurrence without subtractions in indices:
$$
a_{n + 2} = - 4 a_{n + 1} - 4 a_n
$$
Multiply by $z^n$, add over $n \ge 0$, and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = - 4 \frac{A(z) - a_0}{z} - 4 A(z)
$$
By running the recurrence backwards, you have $a_0 = -1$, and:
$$
A(z) = \frac{2}{(1 + 2 z)^2} - \frac{3}{1 + 2 z}
$$
Remember the generalized binomial theorem, in particular for $m \in \mathbb{N}$:
$$
(1 + u)^{-m}
  = \sum_{k \ge 0} \binom{-m}{k} u^k
  = \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
$$
and you can read off:
\begin{align}
a_n
  &= 2 \cdot (-2)^n \binom{n + 2 - 1}{2 - 1} - 3 \cdot (-2)^n \\
  &= (2 n - 1) \cdot (-2)^n
\end{align}
