Intuitive explanation for the second solution. Projectile problem Take the classic problem of a person standing on a high building/cliff and throws an object which follows a parabolic path to the ground.
Solving this problem gives two solutions for the time in which the object will strike the ground. One is positive and one is negative. I can see where this negative solution comes from intuitively. If the person has fired the object in the other direction along the parabola then that is the time it would take to reach ground level.
Now take the problem of a stone dropped in a well. The sound is heard 2.40 s later and the speed of sound in air in the well is 336 m/s. How deep is the well?
My solutions are 26.4 m and 24600 m.
Obviously the first solution is the correct one but where does the second solution come from?
 A: In the building example, you are following the stone on its parabolic trajectory for all time. The negative time solution is the 1st time it crosses $h=0$, reaching the top of the cliff at $t=0$, and then falling as per the problem.
Onto the stone and the well:
If the stone fall a distance $x$ in time $t_0$, then the sound is heard at:
$$ t_1 = \frac x c + t_0 $$
Of course $d$ and $t$ are related by
$$ x = \frac 1 2 g t_0^2 $$
so
$$ t_1 = \frac x c + \sqrt{2x/g} $$
The positive square root gives $x=2.4\,$m, while:
$$ t_1 = \frac x c - \sqrt{2x/g} $$
yields $x=24$,$600\,$m.
That means you are using
$$t_0 = - \sqrt{2x/g} = -70.8\,{\rm s}$$
Again, the stone is on a parabolic trajectory. It starts at the bottom the 24,600m well. The sound is racing upward at $c$, while the stone starts off at:
$$ gt_0 = -694\,{\rm m/s}$$
(down is positive). In the 70-ish second ballistic trajectory, the rock is slowing down, and the sound is catching up. At $t=0$, the stone reaches to top of the trajectory. 2.4 seconds latter, the sound is heard.
