Let $d$ be the depth of the well in feet, let $t_1$ be the number of seconds the rock took to fall to the bottom of the well, and let $t_2$ be the number of seconds the sound took to travel back up the well. We know $t_1+t_2=7.7$. Now we can solve $t_1$ for $d$:
$$d=16t_1^2$$
$$\frac{d}{16}=t_1^2$$
$$t_1=\frac{\sqrt{d}}4$$
And similarly $t_2$:
$$d=1120t_2$$
$$t_2=\frac{d}{1120}$$
So $\frac{d}{1120}+\frac{\sqrt{d}}4-7.7=0$. If we let $u=\sqrt{d}$, this becomes a quadratic.
$$\frac{u^2}{1120}+\frac{u}4-7.7=0$$
$$u=\frac{-\frac14\pm\sqrt{\left(\frac14\right)^2-4(\frac1{1120})(-7.7)}}{\frac2{1120}}$$
$$u=560\cdot\left(-\frac14\pm\sqrt{\frac1{16}+\frac{7.7}{280}}\right)$$
$$u=560\cdot\left(-\frac14\pm\sqrt{\frac{36}{400}}\right)$$
$$u=560\cdot\left(-\frac14\pm\frac3{10}\right)$$
We know $u$ is the positive square root of $d$, so we can replace the $\pm$ with a $+$.
$$u=560\cdot\left(\frac1{20}\right)$$
$$u=\frac{560}{20}$$
$$u=28$$
Then $d=28^2=784$, and the answer is $\boxed{\textbf{(A)}}$.