What is the Maximum Horizontal Distance for a Gazilcher Launch?

[xSPARX]

Homework Statement

Some students at Rice University have built their first
gazilcher – a large slingshot (operated by five people
at once, and capable of launching objects well over 100
meters in distance) constructed from surgical tubing
and a cafeteria silverware basket. They wish to launch
a water balloon through a window on the first story of
a dormitory. The window has a vertical height h relative
to the ground where the gazilcher is activated. In
order to avoid getting caught, the culprits wish to be
as far away as possible. Furthermore, they always aim
their gazilcher for its maximum range, R, which they
have exactly measured. Show that the maximum horizontal
distance between the gazilcher and the target
window, neglecting air resistance, is given by

Homework Equations

d= R/2(1+ √(1-4h/R))

The Attempt at a Solution

i used the angle 90 because it was the maximum range/height regarding that they needed maximum length. I used the equation for Range to get the Height that they had. I seem to not understand how they got the 1-4h/r within a radical.

 

[Dekoy]

Here's the way I did it
1. We know that 45 degrees gives max height and max range.
2. Using projectile motion equation y=xtanB-(1/2)gx^2(VocosB)^-2 set it equal to 0.
3. Since max R=(Vo^2)/g solve for (Vo^2) and substitute.
4. Solve for X after substituting and you're done.

If you need any more help just let me know.