Where is my error in trying to find Pythagorean triples with matching areas?

I notice that the equation is of the following form (I replaced $n$ by $x$):

$x^3 - m^2 x + \frac{D}{m} = 0$

Please see the following reference for the trigonometric solution. This is applicable to your case since the coefficient of $x$ is negative. This means your equation has three real roots.

Trigonometric solution for three real roots

In summary:

Let

$p = -m^2$

$q = \frac{D}{m}$

$a = \frac{-p}{3}$

$b = \frac{q}{2}$

$C(p, q) = 2 \sqrt{a} \cos\left(\frac{1}{3} \arccos \left(\frac{-b}{a} \sqrt{\frac{1}{a}}\right)\right)$

Then the three roots are given by

$t_0 = C(p, q)$

$t_1 = C(p, -q) - C(p, q)$

$t_2 = -C(p, -q)$

I checked that the method works for both of your examples.

Let me include the output of a program I wrote:

For (6, 2)

p:-4.0

q:3.0

Answer:

$t_0$: 1.3027756377319946

$t_1$: 1.0000000000000002

$t_2$: -2.302775637731995

For (2730, 10)

p:-100.0

q:273.0

Answer:

$t_0$: 8.156603957913985

$t_1$: 2.9999999999999982

$t_2$: -11.156603957913983

For (2730, 14)

p:-196.0

q:195.0

Answer:

$t_0$: 13.47318861248212

$t_1$: 1.0

$t_2$: -14.47318861248212