Trying to solve an odd equation minimum. Here is a math puzzle I've been working on -

The set-up - I am a dog, 50 feet from the water. A rubber duck is in the water 50 feet from shore and 140 feet to my right. This makes a triangle base of 140 ft, height 100 ft. I can run 30fps on land, 4fps in water. What is the fastest I can get to the duck?
Given the only real choice is to vary the point he hits the water, I have the image showing that point, and how it produces two right triangles. So we have this equation to minimize. 
$y=\frac{\sqrt{\left(2500+x^2\right)}}{30}+\frac{\sqrt{\left(2500+\left(140-x\right)^2\right)}}{4}$
Graphing this produces the conclusion that at X= 133.71, Y= 17.357. I got this far, but manipulating to get rid of radicals sent me down a rabbit hole. 
Disclosure, a high school junior came to me with this problem. If the solution requires calculus, I'd love to see the method, but the students that had this problem were not that advanced.  
(I'd welcome a better title for this problem)
 A: The $x$-value at which $y$ attains its minimum turns out to be the smaller of the two real roots of the polynomial $11025000000 - 157500000 x + 4884100 x^2 - 61880 x^3 + 221 x^4$, which is irreducible over the integers. So I doubt any by-hand method is going to be very clean.
A: Refer to the below diagram (hopefully this diagram is clear enough, sorry for the poor quality):

We want to reach from the point $P$ to the point $Q$ in the shortest amount of time. For your problem, $a=50,\,b=50$ and $c=140$. Our speed in water $v_w=4$ and on land, $v_l=30$.
Let the optimal point at which we should reach the interface between land and water be at a horizontal distance $x$ from $Q$ (in the direction of $P$).
The time taken to traverse $PR$ is $\dfrac{\sqrt{(c-x)^2+b^2}}{v_l}$ and to traverse $RQ$ is $\dfrac{\sqrt{a^2+x^2}}{v_w}$.
So, we want to minimize $t=\dfrac{\sqrt{(c-x)^2+b^2}}{v_l}+\dfrac{\sqrt{a^2+x^2}}{v_w}$.
We differentiate the above with respect to $x$ and set it equal to $0$.
We get, $\dfrac{-(c-x)}{v_l\sqrt{(c-x)^2+b^2}}+\dfrac{x}{v_w\sqrt{a^2+x^2}}=0$
or $\dfrac{x}{v_w\sqrt{a^2+x^2}}=\dfrac{(c-x)}{v_l\sqrt{(c-x)^2+b^2}}$.
From the figure, $\dfrac{x}{\sqrt{a^2+x^2}}=\sin m$ and $\dfrac{(c-x)}{\sqrt{(c-x)^2+b^2}}=\sin n$.
Thus, $\dfrac{\sin m}{v_w}=\dfrac{\sin n}{v_l}$.
Hence, we want to reach the water at a point such that the angles $m$ and $n$ satisfy the above equation. As is clear from the geometry, the angles $n$ and $m$ are constrained such that if one of them is determined then both of them are. We can obtain this constraint by noting that $x=c-b\tan n$ and thus, $\tan m=\dfrac{x}{a}=\dfrac{c-b\tan n}{a}$.
This is not a complete solution since solving these two equations for $m$ and $n$ is probably as difficult as solving the quartic presented by Greg Martin. But this hopefully provides a nice interpretation to the nature of the solution.

This problem is usually discussed in relation to Snell's law in ray optics. Snell's law can be derived from Fermat's principle which states that the path taken by light between two points is such that it takes light the least amount of time to traverse it. Thus, you can interpret the above problem as a ray of light going from point $P$ to $Q$ with speed $v_l$ in air and $v_w$ in water. Since light follows the principle of least time, we have shown the above relation $\dfrac{\sin m}{v_w}=\dfrac{\sin n}{v_l}$ (which is Snell's law) holds for light (go to the links for more details).
