Going off of @lioness99a's comment, for your example you can set $a=2$, $b=5y$, and $c=-8y^2$ to get
\begin{align*} x &= \frac{-5y \pm \sqrt{(5y)^2 - 4\cdot 2 \cdot (-8y^2)}}{2\cdot 2} \\
&= \frac{-5y \pm \sqrt{89}|y|}{4} \\
&= \frac{1}{4} (-5 \pm \sqrt{89})y
\end{align*}
or fix $x$ and treat $y$ as the variable.
AFAICT the trig method for solving low degree polynomial relies on the properties of Chebyshev polynomials.
Another starting point is that the solutions of
$$
\cos n\alpha=\cos \beta
$$
are
$$
\alpha=\frac1n(\pm \beta+\frac k{2\pi})\tag{1}
$$
with $k\in\Bbb{Z}$. This is an immediate consequence of $\cos\alpha=\cos\beta\Leftrightarrow \alpha=\pm\beta+k\cdot2\pi$.
Chebyshev polynomials come into play via the relation
$$
T_n(\cos x)=\cos nx.
$$
Let's start with $n=3$. Then $T_3(t)=4t^3-3t$, all according to the familiar triplication formula $$\cos 3\alpha=4\cos^3\alpha-3\cos\alpha.$$
If you can write a cubic equation in the form
$$4x^3-3x=y,\tag{2}$$
then the substitutions $x=\cos\alpha$, $y=\cos\beta$ turn $(2)$ into
$\cos3\alpha=\cos\beta$, and we are in business. It follows that
$$
x=\cos\left(\pm\frac13\left(\arccos y+k\cdot2\pi\right)\right).
$$
Because cosine is even, we can ignore $\pm$. Because cosine has period $2\pi$, it suffices to only include $k=0,1,2$.
The upshot is that a general cubic can be brought into the form $(2)$ by a linear subsititution. You first need to depress the quadratic term, and then you need to arrange the coefficients of the cubic and the linear terms have that $4:(-3)$ ratio.
For example, with your cubic
$$m^3n-mn^3=D$$
we easily see that the substitution $n=2mx/\sqrt3$ brings it to the form
$$
\frac{8m^4}{3\sqrt3}x^3-\frac{2m^4}{\sqrt3}=D,
$$
which is a scalar multiple of
$$
4x^3-3x=\frac{3\sqrt3 D}{2m^4}.
$$
When $n=2$ we have $T_2(t)=2t^2-1$ meaning that the solutions of
$$
2x^2-1=y\tag{3}
$$
are
$$
x=\cos\left(\frac12\left(\arccos y+k\cdot2\pi\right)\right)
$$
with $k=0,1$. Again, a linear substitution brings a general quadratic to the form $(2)$, first be depressing the linear term, and then scaling the variable linearly. In the case of a quadratic this is not usually done, because we have the simple quadratic formula.
Unfortunately with $n>3$ we lose the key ability to bring a general polynomial into the desired form involving a Chebyshev polynomial. We can find the solutions to a quartic of the form
$$
8x^4-8x^2+1=y
$$
as
$$
x=\cos\left(\frac14\left(\arccos y+k\cdot2\pi\right)\right)
$$
with $k=0,1,2,3$, but the trickery with linear substitutions does not work (at least not alone). We can depress the cubic term, and fix the ratio of the coefficients of the quadratic and quartic terms, but we cannot kill the linear term. With $n=5$ and higher it becomes worse.
Best Answer
Consider the quadratic $ax^2 + bx + c$. If $a = 0$, then finding the root is easy and boring, so we only look at the case when $a \neq 0$.
One way to get at the original quadratic formula is to complete the square. That is, note that $$\begin{align} ax^2 + bx + c &= a\left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) \\ &= a \left(x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a} \right) \\ &= a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right). \end{align}$$ So when looking for solutions to $ax^2 + bx + c = 0$, we are looking at solutions to $$ 0 = a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right) = a\left( x + \frac{b}{2a}\right)^2 + a\frac{4ac - b^2}{4a^2},$$ or equivalently $$ \left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.$$ Taking square roots and solving for $x$ gives the original quadratic formula.
Now let us derive the alternate form. Notice that if $c = 0$ then $x = 0$ is a root and $x = -b/a$ is the other root. So let us now consider $c \neq 0$. We now know that $x = 0$ is not a root. This means that if $r$ is a root of $ax^2 + bx + c = 0$, then $\frac{1}{r}$ will be a root of $a \frac{1}{x^2} + b \frac{1}{x} + c = 0$. Multiplying through by $x^2$ (which makes sense because we are not interested in $x = 0$), we are interested in reciprocals of the roots of $$ a + bx + cx^2 = 0.$$ The original quadratic formula applies, and we know the roots are $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}.$$ Since we are interested in the reciprocals, we find the roots to be $$ x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}.$$
This isn't necessarily the best derivation, but I like it. $\diamondsuit$