One of the easiest ways to do problem one is to factor it, which it already is you have p and (120-4p). This means that p = 0 and 120-4p = 0. If you algebraically solve for p, you will find two answers 0 and 30 tickets sold makes $0 in revenue. For number two we know that 120p-4p^2 is equal to the revenue. So to create an equation that solves for the number of tickets to make $600 we make the equation 120p-4p^2 = 600, but because we want to get rid of the 600 on the right, we subtract 600 from both sides to get 120p-4p^2-600
