'66 covered it great. If your meter shows a Negative "-" sign in front of the DC voltage, that would indicate the leads are connected in reverse polarity. Switch the leads around and there should only be the DC voltage shown on the display. Observing the marked polarity on the meter's plugs will show which polarity is on the Battery [or even a cell]. "+" on the meter's probe plug will equal up with the "+" polarity on the Battery, or Cell. Same goes with the "-" polarity.
Definitely charge the Battery - or Cell - in Parallel, meaning the "+" polarity from the charger connects to the "+" polarity on the Battery / Cell. Same follows for the "-" polarity.
If the Battery you have is a Liquid Battery, the normal Voltage at full charge and no-load should be no less than 6.6 VDC. Try to charge it with a charger that has a loaded output of at least 7.0 VDC or maybe even 8.0 VDC. Since it has been fully discharged, you should trickle charge it for a few hours, then charge it at a higher level until it is near full charge, then let it trickle charge over night [I will re-check this procedure for lead-acid Batteries being recharged from a dead or 1.0 spec gravity state].
If it's a NiCAD, it might have reversed it's own polarity from being fully discharged and left at that state for an extended period [they are funny that way
..]. If the Battery is NiCAD, the voltage under load, upto the point where it dies, will remain a steady 6.0 VDC [5 - 1.2 VDC cells]. Open circuit, or no-load voltage might be as high as 10 VDC, which is why chargers for NiCADs have such a high charging voltage.
If you can charge the NiCAD, it should be done with a fast charge [prevents "memory" oxidation]. Since it is dead, the battery will draw an excessive amount of current for the first 10 to 15 minutes, before tapering off to a somewhat steady value. This might overload the charger, so keep this in mind.
On the subject of reversed polarity charger connections, hooking up a dead battery to a charger with reversed polarity will most likely fry the charger, since it will eventually become a series additive circuit, doubling the voltage across a low resistance load, which now will draw twice the current as before [once again, I need to check my books on this one, as it's been a few years since I have refreshed the subject].
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