(20-0.3n) – 2

Total Profit:

n((20-0.3n)-2)

Formula (if he wants to make 240 profit):

n((20-0.3n)-2) = 240

n(18-0.3n) = 240

18n – 0.3n^2 = 240

(multiply both sides by 10/3)

60n – n^2 = 800

-n^2 + 60n – 800 = 0

(n – 20) x (- n + 40) =0

n= 20, 40

So, the retailer will make a profit if he sells between or at 20-40 games. This is a quadratic equation, so, you’ll see that his profit is above or at 240 if he sells between 20 and 40 games.

n(20 – 0.3n) – 2n = 240

20n – 0.3n^2 – 2n = 240

18n – 0.3n^2 = 240

for simplicity I would divide everything by -0.3

n^2 – 60n = -800

n^2 – 60n + 800 = 0

now factor

(n – 20)(n – 40) = 0

so the problem has 2 solutions n = 20 or n = 40

check the solution with the original equation

20(20 – 0.3(20)) – 2(20) = 240 (true)

40(20 – 0.3(40)) – 2(40) = 240 (true)

both answers check so the solution is n = 20 or 40

Cost of goods = 2x

Selling price = (20-0.3n)x

Profit = 20x-0.3nx-2x = 240

18x – 0.3nx = 240

x = 240/(18-0.3n) … in terms of ‘n’