Turing-decidable
A Turing-decidable language is one some machine handles like an impeccable, never-stuck bouncer: shown any ID, member or not, he always reaches a firm decision in finite time and says yes or no. No staring forever, no hanging, no 'come back later'. Every string gets a clear, correct verdict, and the machine always stops to give it.
Formally, a language L is Turing-decidable (also called recursive) if there is a Turing machine M that halts on EVERY input and accepts exactly the strings in L: it halts in q_accept for strings in L and halts in q_reject for strings not in L. The two requirements together are the point: correctness (it accepts members and rejects non-members) AND the guarantee to halt on all inputs, never looping. Such a machine is called a decider. Deciding a language is the precise formalization of having an algorithm that always terminates and always answers the membership question correctly.
Decidability is the gold standard: a decidable problem is one you can actually solve by a guaranteed-to-finish algorithm. It sits strictly inside Turing-recognizable: every decidable language is recognizable (a decider is in particular a recognizer), but not vice versa, because a recognizer may loop on non-members. A clean theorem ties them together: a language is decidable exactly when both it AND its complement are Turing-recognizable. That is why showing the halting problem's complement is not recognizable proves the halting problem is undecidable.
The language {a^n b^n c^n : n >= 0} is Turing-decidable: a machine marks one a, one b, one c per pass, halts-and-accepts when all are matched evenly, and halts-and-rejects on any mismatch or wrong order. It always stops, on every input. (Note this language is not even context-free.)
Decidable: the machine always halts and always gives the right yes/no answer.
Decidable requires halting on EVERY input, not just members. A language is decidable exactly when both it and its complement are recognizable; this is the lever for proving undecidability.