Dropping a coin and finding out the depth

Question:
You released a coin in order to find the depth of a well. The time between dropping the coin and hearing it hit the bottom is 2.059 s. The speed of sound is $v_s=$343 m/s. Calculate the depth, $d$, of the well.

Answer:
The total time $T$ is divided into the time hitting the bottom, $t_1$ and time the sound traveling, $t_2$. Namely,
\[T=t_1+t_2\]
We have
\[ d = \frac{1}{2}gt_1^2 \\
  t_2 = \frac{d}{v_s}\]
Since $t_1 = T - t_2$, the depth can be expressed as
\[ d = \frac{1}{2}g(T-\frac{d}{v_s})^2 \]
Expand and arrange it in terms of $d$.
\[ d^2 - d(\frac{2v^2}{g}+2v_sT)+v_s^2T^2=0 \]
Plug $v_s=$343 m/s and $g=$9.8 m/s$^2$ into the above.
\[ d^2 -25,422.5d + 498,770.7 = 0 \]
The solution is
\[ d=\frac{-(25,422.5)\pm\sqrt{(25,422.5)^2-4\times 498,770.7}}{2} \\
   = 19.6 \mathrm{m} \]
The other solution is neglected.

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