Laura is putting together the following list: $a_0, a_1, a_2, a_3, a_4, ..., a_n$, where $a_0 = 3$ and $a_1 = 4$.
She knows that the following equality holds for any value of $n$ integer greater than or equal to $1$:
$$a_n^2-2a_{n-1}a_{n+1} =(-2)^n.$$Laura calculates the value of $a_4$. What value does it get?
1. We start with the given sequence \(a_0, a_1, a_2, \ldots, a_n\) where \(a_0 = 3\) and \(a_1 = 4\). The recurrence relation is given by:
   \[
   a_n^2 - 2a_{n-1}a_{n+1} = (-2)^n
   \]

2. To find \(a_2\), we use the recurrence relation for \(n = 1\):
   \[
   a_1^2 - 2a_0a_2 = (-2)^1
   \]
   Substituting the known values \(a_0 = 3\) and \(a_1 = 4\):
   \[
   4^2 - 2 \cdot 3 \cdot a_2 = -2
   \]
   Simplifying:
   \[
   16 - 6a_2 = -2
   \]
   Solving for \(a_2\):
   \[
   6a_2 = 18 \implies a_2 = 3
   \]

3. Next, we find \(a_3\) using the recurrence relation for \(n = 2\):
   \[
   a_2^2 - 2a_1a_3 = (-2)^2
   \]
   Substituting the known values \(a_1 = 4\) and \(a_2 = 3\):
   \[
   3^2 - 2 \cdot 4 \cdot a_3 = 4
   \]
   Simplifying:
   \[
   9 - 8a_3 = 4
   \]
   Solving for \(a_3\):
   \[
   8a_3 = 5 \implies a_3 = \frac{5}{8}
   \]

4. Finally, we find \(a_4\) using the recurrence relation for \(n = 3\):
   \[
   a_3^2 - 2a_2a_4 = (-2)^3
   \]
   Substituting the known values \(a_2 = 3\) and \(a_3 = \frac{5}{8}\):
   \[
   \left(\frac{5}{8}\right)^2 - 2 \cdot 3 \cdot a_4 = -8
   \]
   Simplifying:
   \[
   \frac{25}{64} - 6a_4 = -8
   \]
   Solving for \(a_4\):
   \[
   6a_4 = \frac{25}{64} + 8 = \frac{25}{64} + \frac{512}{64} = \frac{537}{64}
   \]
   \[
   a_4 = \frac{537}{64 \cdot 6} = \frac{537}{384} = \frac{179}{128}
   \]

The final answer is \(\boxed{\frac{179}{128}}\)