$$\frac{ab^2}{x^2} = \frac{c (P-Q) x^3}{P}$$
I've got the expression above and should express it in terms of $x$, knowing that $a, b, c$ and $Q$ are constants and $P = \sqrt{Q^2 + \frac{x^2}{4}}$
Also, we know that $Q >> x$, so I decided to use the first terms of the Taylor expansion for $P$, so $$P \approx Q(1+\frac{x^2}{8Q^2})$$
By doing so, we get $ab^2(Q + \frac{x^2}{8Q}) = c(\frac{x^2}{8Q})x^5$
In order to solve for $x$, I would have to solve this $7^{th}$ degree equation. So, assuming that $x^2 \approx 2x$, we get a $5^{th}$ degree equation
$ab^2(Q + \frac{x}{4Q}) = c(\frac{x}{4Q})x^5$ wich looks less ugly, but I still can't solve it.
Do you guys have any ideas?
I thank you in advance for your cooperation.
multiplying both sides by $x^2$ we get $$ab^2=\frac{c(P-Q)x^5}{P}$$ and then multiply by $$\frac{P}{c(P-Q)}$$ you will get $$\frac{ab^2P}{c(P-Q)}=x^5$$ $$x=\sqrt[5]{\frac{ab^2P}{c\left(P-\sqrt{D^2-\frac{x^2}{4}}\right)}}$$ and now you will Need a numerical method! you can not solve it by an explicit formula, it is an higher order polynomial with your correction you will get $$ab^2(4Q^2+x)=cx^6$$ and for this equation you will also Need a numerical method!