Solving a simple rational equation $(\frac{6x}{6-x})^2+x^2=400$ Clearly we could multiply both sides of
$$\left(\dfrac{6x}{6-x}\right)^2+x^2=400$$
by $(6-x)^2$ which leads to a degree 4 polynomial equation, which we can solve using the bi-quadratic formula. Moreover, we could approximate the solutions using Newtons method.  However, I have a feeling there is a much more graceful way to solve this. Any help is appreciated. 

The quartic (or bi-quadratic) polynomial for $x$ can be factored into two quadratic polynomials with coefficients in $\mathbb{Z}[\sqrt{109}]$, so that the closed-form expression for roots is not as terrible as it can be generally.
Clearing fractions one gets:
$$ x^4 - 12x^3 - 328x^2 + 4800x - 14400 = 0 $$
Then notice that:
$$ x^4 - 12x^3 - 328x^2 + 4800x - 14400 = (x^2 - Ax + 6A)(x^2 - Bx + 6B) $$
provided $A+B = 12$ and $AB = -400$.
It follows that the roots for $x$ may be found by solving the respective quadratic equations with $A = 6 + 2\sqrt{109}$ and $B = 6 - 2\sqrt{109}$.
More explicitly:
$$ x = 3 - \sqrt{109} \pm \sqrt{82 + 6\sqrt{109}}\;,\; 3 + \sqrt{109} \pm \sqrt{82 - 6\sqrt{109}} $$
A crude estimate shows that the expressions under the radicals are all positive, and thus the four roots are indeed real roots.
How was the reducibility of the polynomial recognized?
Initially I saw that setting $y = \frac{6x}{x-6}$ leads to a system of equations:
$$ x^2 + y^2 = 20^2 $$
$$ (x - 6)(y - 6) = 36 $$
Geometrically this asks for the intersection of a circle of radius $20$ centered at the origin and a "right" hyperbola with center $(6,6)$.  Since both curves are symmetric about the line $y = x$, it follows that when $(x,y)$ is a point of intersection, so too is $(y,x)$.
Algebraically this amounts to saying that $x \mapsto \frac{6x}{x-6}$ is an involution which pairs up the roots $x$.  A bit of scratchwork showed that the pairing of roots in this way results in factors $x^2 - Ax + 6A$ and $x^2 - Bx + 6B$ where $A,B$ are roots of $w^2 - 12w - 400 = 0$.
