What are you trying to do?

You have a demand function:

where

denotes the quantity that can be sold (the demand) at unit price

. Then the revenue is:

Now there are several methods of finding the price

that maximises the revenue

. If you know calculus you can differentiate the revenue with respect to price, set the derivative to zero and solve for the price which will be the price that maximises revenue... Alternatively you can complete the square (which if you do not have calculus you will have been taught either in your current course of your algebra or pre-calculus course).

Either way you will find the price which maximises revenue is 60 quatloos from which you can find the demand and revenue at the maximum.

CB