If you set a particle to move in a frictionless bowl (radius R) at the velocity v, how far is the orbit from the equator?

I got the following equations:

[tex]tan\alpha = \frac{v^2}{gr}[/tex]

[tex]r = Rsin\alpha[/tex]

[tex]d = Rcos\alpha[/tex]

where g is the constant 9,81[tex]\frac{m}{s^2}[/tex] and r the radius of the orbit.

I ended up with [tex]d = \frac{\frac{v^2}{gR}-\sqrt{(\frac{v^2}{gR})^2+4}}{-2}R[/tex], which seems a little complicated but gives reasonable answers. I haven't yet figured out why it only works with - in front of the square root (I got both plus and minus when I solved the equations). Is my solution correct?