數學~二次方程

一名商人以$18000的價錢購入一批恤衫，然後他將該批恤衫全部出售，除了其中的5件恤衫未能售出外，其餘售出的恤衫每件都能替他帶來$20的利潤。該名商人將售出該批恤衫而得到的全部款項(包括利潤)再用來購入另外一批成本價完全一樣的恤衫，但他這次可以多購入30件恤衫，試求每件恤衫的成本價。
設原本購入恤衫數是x﹐則新購入恤衫數是x+30, 另設每件恤衫的成本價y
恤衫每件能替商人帶來$20的利潤﹐售價因而是y+20
則
xy=18000...(1) [恤衫數*每件恤衫的成本價=總成本
又該名商人售出該批恤衫而得到的全部款項
=(x-5)(y+20) [因有5件未能賣出]
用這全部款項﹐商人可以多購入30件恤衫
(x-5)(y+20)=(x+30)y
xy+20x-5y-100=xy+30y
20x=35y+100
y=(4/7)x-(20/7)
代入
xy=18000
x[(4/7)x-(20/7)]=18000
x[(4x-20)]=126000
4x^2-20x-126000=0
x^2-5x-31500=0
x=180 or -175 (捨去) [恤衫數目不能是負數]
y=(4/7)x-(20/7)=100
原本購入恤衫數是180件﹐每件恤衫的成本價是100元
新購入恤衫數是210件
希望幫到你啦

Let the cost of each 恤衫 be $x
number of 恤衫 in first round = 18000/x
number of 恤衫 sold = 18000/x - 5
利潤 = (18000/x - 5)*20
Total amount get from first selling = (18000/x - 5)*(20+x)
Total cost in second round = (18000/x + 30)x
Thus, (18000/x + 30)x = (18000/x - 5)*(20+x)
18000+30x = 360000/x - 100 + 18000 - 5x
35x + 100 = 360000/x
35x^2 + 100x - 360000 = 0
7x^2 + 20x - 72000 = 0
By solving, x = 100 or -102.6/7(rej)
Thus, cost =$100

Let Y be the number of shirts purchased in his FIRST purchase.
Then, the cost of each shirt = $18000 / Y

Amount received from sales
　= (Y-5) (18000/Y + 20)
Amount spent on making his SECOND purchase
　= (Y+30) (18000/Y)
So,
(Y-5) (18000/Y + 20) = (Y+30) (18000/Y)
35 (18000/Y) = 20 (Y-5)
35 (900/Y) = Y-5
Y² - 5Y - 31500 = 0
(Y - 5/2)² - 126025/4 = 0
Y = 5/2 ± 355/2
　= 180 or -175 (rejected as Y＞0)
Y = 180
Therefore, cost of each shirt = $18000 / 180 = $100