It's a physics-related problem, but it has a nasty equation:
Let the speed of sound be $340\dfrac{m}{s}$, then let a heavy stone fall into the well. How deep is the well when you hear the impact after $2$ seconds?
The formula for the time it takes the stone to fall and the subsequent sound of impact to travel upwards is simple enough:
$t = \sqrt{\dfrac{2s}{g}} + \dfrac{s}{v}$
for $s$ = distance, $g$ = local gravity and $t$ = time.
This translates to said nasty equation:
$gs^2 - 2sv^2 + 2gstv + gt^2v^2 = 0$
Now I need to solve this in terms of $s$, but I'm at a loss as to how to accomplish this. How to proceed?