What is the oxidation number of the #"Mo"# in #"MoO"_4^(2-)#?
Oxygen almost always has an oxidation number of -2, so it is safe to make the assumption that it does here (the exception are certain peroxides and superoxides).
Four oxygens give a total of (4 x -2) = -8.
The Molybdenum must be positive, but if the complex overall has a charge of -2, there are 2 negative charges left over.
To put it mathematically: