Remember the rule ? well, if you square both sides of the equation, you'll remove the square root.
Then, you can rearrange to remove the fraction, and solving for should be easier.
Almost, if my calculations are correct (this is quite a tedious one!) you should have . You want to try and isolate , start by getting rid of that fraction by multiplying both sides by . Once that's "on the flat", simplify / expand where necessary so you can subtract the term on its own side of the equation and work from there.
Looks like this is the final/accepted version
Let k = SQRT[(1 - r^2)^2 + (2zr)^2] ; then:
x = MRr^2 / (mk)
k = MRr^2 / (mx) ; square both sides:
k^2 = M^2 R^2 r^4 / (m^2 x^2) ; substitute back in:
(1 - r^2)^2 + (2zr)^2 = M^2 R^2 r^4 / (m^2 x^2) ; rearrange:
4 z^2 r^2 = M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2
z^2 = [M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2) ; sooooooooooo:
z = SQRT{[M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2)}
Can be slightly simplified to:
z = SQRT[M^2 R^2 r^4 - (m^2 x^2)(1 - r^2)^2] / (2rmx)