Projectile: Given V0, h, show d=(v0/g)sqrt((v0)^2-4gh)

[qamptr]

Homework Statement

A projectile is fired from a gun (adjusted to give maximum range) with velocity v_{0}. The projectile passes through two points at a height h. The problem asks us to show that d=\frac{v_{0}}{g}\sqrt{v^{2}_{0}-4gh}
where d is the distance between the two points at height h.

Homework Equations

r=v_{0}t+\frac{1}{2}at^{2}
v=v_{0}+at
x= \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

The Attempt at a Solution

I was able to get a quadratic function of x:
0=\frac{g}{v^{2}_{0}}x^{2}-x+h

After manipulation using the quadratic formula, all I can see is:
x=\frac{v^{2}_{0}}{2g}+\frac{1}{v_{0}}\sqrt{v^{2}_{0}-4gh}

Which just looks so close but I'm killing myself in trying to see how it is either (1) wrong or (2) able to be simplified.

EDIT: x=\frac{v^{2}_{0}}{2g}+\frac{1}{2gv_{0}}\sqrt{v^{2}_{0}-4gh}, sorry.

Help?

 

[qamptr]

Does anyone even have any suggestions? This is actually due in about an hour and a half. I'm not heartbroken or anything but I'm feeling pretty annoyed that I might not get this problem. I honestly can't see what's going wrong here. Any creative suggestions or strong nudges are totally welcome.

Thanks...

 

[qamptr]

Alright, so, I currently have the following written on my paper:

x= \frac{v^{2}_{0} \pm v_{0} \sqrt{v^{2}_{0}-4gh}}{2g}

I can't find the correction that makes this into the formula asked for.

 

[qamptr]

Ahhhhhhh... So, this one was staring me in the face. Done and with 20 minutes to spare.