tl;dr: most likely not solvable with the information given.
The next number is $-642$ and the pattern is formed by
$\frac{1}{18}\cdot{}x^7
- \frac{1411}{720}\cdot{}x^6
+ \frac{19441}{720}\cdot{}x^5
- \frac{26609}{144}\cdot{}x^4
+ \frac{94175}{144}\cdot{}x^3
- \frac{49714}{45}\cdot{}x^2
+ \frac{12509}{20}\cdot{}x^1
+177$.
Well, probably not: it isn't stated in the question, but the solution presumably is a positive integer (though regression analysis suggests that it has to be negative).
The most straightforward solution actually is simple addition:
$510 = 192+30+55+86+43+22+79+3$
More serious considerations:
I shall be assuming that this is purely mathematical, i.e. not relying on trivia like the character's names, the reciptionist's birthday or the arrangement of buttons on the input pad. Furthermore people go into finance because they are bad at maths, so only elementary operations are permitted and the numbers are used in the order they show up in (and not reverse or sorted ascending or descending, alternating front/back or some non-derivable order). Nor shall there be skips in the sequence (e.g. taking only the prime-numbered elements of the sequence generated).
If I were Lalozik I would have the sequence of numbers change every time the safe is attempted to be opened: while the rule for generation and solution would remain the same, the seed for generation would change randomly, hence one couldn't predict the next number without knowing the algorithm. However, this is obviously not the case, since otherwise Banes would have staked out the bank beforehand and seen the patterns and possibly the solutions to previous times the safe was opened, so those sequences would be provided to us for solution.
For it to be possible to derive a solution by looking at the list of numbers with such restricted information, the algorithm has to repeat: for example, for $1, 2, 3, 5, 8$ one would guess the fibonacci numbers where $x_{n+2}=x_{n+1}+x_{n}$, but this cannot be derived if only $1, 2$ are given since $3$ would be the very first number where the formula applies. Indeed, even $1, 2, 3$ doesn't suffice: there need at least two repeats of the application of the formula within the sequence, one to establish the pattern and one to affirm it.
Otherwise if someone came up with a different solution one couldn't dismiss it: if there is no repeating pattern then straightforward addition is the simplest solution. Hence at most 7 of the numbers can be used to derive the solution, the 8th must serve for confirmation - the fewer needed the more plausible the rule becomes.
In fact, it can be easily shown (google "common difference") that in the present case there exists no polynomial that can generate these numbers in a manner that permits the solution to be confirmed. Of course, this still allows any algorithm that is more complex than a polynomial (e.g. involving floor functions or cosines) - though then the claim that the scheme is "very simple and straightforward" is misleading at best.