PS....

On Feb 15, 7:49*pm, I wrote:
Here is the formula set up to solve for MGD (Million Gallons per Day).
=3*936*A21^1.5*1440/1000000
[....]
What I want to do is set up another formula where the desired MGD
could be entered in a cell and the result would be the inches to let
flow over the weir.

[...] is this simple algebra? *If, say, B21 is the MGD and you want A21
to be the inches going over the weir, then:

A21^1.5 = B21*1000000/3/936/1440

I certainly prefer Rick's approach of taking 2/3 power of both sides.
Not sure why he has two variables, where I have one. Perhaps Rick has
given this more thought than I have. But I would say:

A21: =(B21*1000000/3/936/1440) ^ (2/3)

Then I was going to point out a simplification of my own formula,
which I will now apply to the one above. That is equivalent to:

=B21^(2/3) * (1000000/3/936/1440)^(2/3)

The advantage of that is: (1000000/3/936/1440)^(2/3) can be computed
one time in some cell, say A1. Then we have:

A21: =A1 * B21^(2/3)

Again, check by writing in C21:

=3*936*A21^1.5*1440/1000000