# Quadratic equation related to physics problem - how to proceed?

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?

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You know how to solve $As^2+Bs+C=0$, so what are $A$, $B$ and $C$ here? – Did Jun 2 '12 at 12:43
What is nasty about it? It is just an ordinary quadratic equation in $s$. I grant you that the coefficients are a bit messy, but that shouldn't stop you for plugging them into the standard solution for quadratic equations. (Do join the two middle terms into $2(gtv-v^2)s$ before you proceed, of course.) – Harald Hanche-Olsen Jun 2 '12 at 12:43
Oh? You have $g=10$, $v=340$ and $t=2$ so you can solve the equation for $s$. How is it nasty? I'd get offended if I were the equation! – Gigili Jun 2 '12 at 12:43
I'm quite ashamed to admit that I kind of forgot to plug in the numbers. – Miroslav Cetojevic Jun 2 '12 at 16:23

Plugging $g$, $t$ and $v$, your equation will be:
$10s^2 - 2(340)^2s + 4 \cdot 10 \cdot 340s + 10 \cdot2^2 (340)^2 = 0$
Hint: if you insert the values of $g, t$ and $v$ you have a quadratic equation in $s$. Even if you just regard $g, t$ and $v$ as constants, you can plug this into the quadratic formula.