QBIC, 31 bytes

~:>:|≈:-a>0|q=q+1┘c=c-a+b]?q\?0

##Explanation

~           IF
 :          cmd line arg 'a'  (the increment of our snail)
  >         is greater than
   :        cmd line arg 'b'  (the decrement, or daily drop)
    |       THEN
≈           WHILE
 :          cmd line arg 'c'  (the height of the well)
  -a        minus the increment (we count down the hieght-to-go)
    >0|     is greater than 0 (ie while we haven't reached the top yet)
q=q+1       Add a day to q (day counter, starts at 1)
┘           (syntactic linebreak)
c=c-a+b     Do the raise-and-drop on the height-to-go
]           WEND
?q          PRINT q (the number of days)
\?0         ELSE (incrementer <= decrementer) print 0 (no solution)

Try it online! (OK, not really: this is a translation of QBIC to QBasic code run in repl.it 's (somewhat lacking) QBasic enviroment)

QBIC, 31 23 bytes

Just noticed the requirements changed. This version doesn't check if the snail will ever reach the top of the well.

≈:-:>0|q=q+1┘a=a-b+:]?q

The explanation below, for the original version that does check if a solution exists, covers all relevant parts of this code too.

Original, 31 byte answer:

~:>:|≈:-a>0|q=q+1┘c=c-a+b]?q\?0

##Explanation

~           IF
 :          cmd line arg 'a'  (the increment of our snail)
  >         is greater than
   :        cmd line arg 'b'  (the decrement, or daily drop)
    |       THEN
≈           WHILE
 :          cmd line arg 'c'  (the height of the well)
  -a        minus the increment (we count down the hieght-to-go)
    >0|     is greater than 0 (ie while we haven't reached the top yet)
q=q+1       Add a day to q (day counter, starts at 1)
┘           (syntactic linebreak)
c=c-a+b     Do the raise-and-drop on the height-to-go
]           WEND
?q          PRINT q (the number of days)
\?0         ELSE (incrementer <= decrementer) print 0 (no solution)

Try it online! (OK, not really: this is a translation of QBIC to QBasic code run in repl.it 's (somewhat lacking) QBasic enviroment)

QBIC, 31 bytes

~:>:|≈:-a>0|q=q+1┘c=c-a+b]?q\?0

##Explanation

~           IF
 :          cmd line arg 'a'  (the increment of our snail)
  >         is greater than
   :        cmd line arg 'b'  (the decrement, or daily drop)
    |       THEN
≈           WHILE
 :          cmd line arg 'c'  (the height of the well)
  -a        minus the increment (we count down the hieght-to-go)
    >0|     is greater than 0 (ie while we haven't reached the top yet)
q=q+1       Add a day to q (day counter, starts at 1)
┘           (syntactic linebreak)
c=c-a+b     Do the raise-and-drop on the height-to-go
]           WEND
?q          PRINT q (the number of days)
\?0         ELSE (incrementer <= decrementer) print 0 (no solution)

QBIC, 31 bytes

~:>:|≈:-a>0|q=q+1┘c=c-a+b]?q\?0

##Explanation

~           IF
 :          cmd line arg 'a'  (the increment of our snail)
  >         is greater than
   :        cmd line arg 'b'  (the decrement, or daily drop)
    |       THEN
≈           WHILE
 :          cmd line arg 'c'  (the height of the well)
  -a        minus the increment (we count down the hieght-to-go)
    >0|     is greater than 0 (ie while we haven't reached the top yet)
q=q+1       Add a day to q (day counter, starts at 1)
┘           (syntactic linebreak)
c=c-a+b     Do the raise-and-drop on the height-to-go
]           WEND
?q          PRINT q (the number of days)
\?0         ELSE (incrementer <= decrementer) print 0 (no solution)

Try it online! (OK, not really: this is a translation of QBIC to QBasic code run in repl.it 's (somewhat lacking) QBasic enviroment)