quadratic equation

Let x be the number of fish May buy originally ,The original cost per fish = $ 600/x ,The totsl no of fish has increased by 1 .May decides to sell all the fish at $ (600/x - 3) per fish , however, she still makes a profit of $27 :(x + 1) (600/x - 3) = 600 + 27(x + 1) (600/x - 3) = 627(x + 1) (600 - 3x) = 627x(x + 1) (200 - x) = 209x200x + 200 - x^2 - x = 209xx^2 + 10x - 200 = 0(x - 10)(x + 20) = 0x = 10 or x = - 20(rejected)