I have two curves. The first is a simple parabola described by
$$y = x^2+bx+c$$
The second is a little more complicated and is described by
$$y = ((n-x^2)/2/x)^2$$
I am trying to find the $x$ and $y$ intersections of those two curves (provided they intersect) but I cannot work out the formula to use.
I tried the trustworthy WolframAlpha and what it returned seemed to be far too complex. Is it really that complex?
Using the input to Wolfram Alpha $$\frac{\left(n-x^2\right)^2}{4 x^2}=x^2+b x+c$$ reduces to $$3 x^4+4 b x^3+2(2 c+n) x^2-n^2=0\tag 1$$ Now, have a look here and enjoy the fun of the analytical solutions of quartic polynomials !
You could better consider numerical methods or, using graphics, consider the intersection of the two functions $$f(x)=3x^2+4bx+2(2c+n) \qquad \qquad g(x)=\frac {n^2}{x^2}$$ This will tell you that there are exactly two real roots.
If, in the Wikipedia page, you look at the pleasant $\Delta$ and apply it to $(1)$, you should get $$\Delta=-2304 n^4 \left(8 c^2+8 c n+5 n^2\right)$$ which is always negative which implies two distinct real roots and two complex conjugate non-real roots.
Considering the behavior of functions $f(x)$ and $g(x)$, you know that one of the roots is positive while the other one is negative.