二次方程 (F.4)

唔知你係唔係咁寫, 應該好似係漏左個括弧
(2x - 5)^2 = (9x - 3)(6x - 15)
(2x - 5)(2x - 5) = (9x - 3)(6x - 15)
4x^2 - 20x + 25 = 54x^2 - 153x + 45
0 = 50x^2 - 133x + 20

唔知可唔可以factor到, 所以用quadratic formula
x = (-b +/- #(b^2 - 4ac) )/(2a) #() = square root (eg. #(4) = 2)
x = (133 +/- #(17689 - 4(50)(20)) )/(100)
自己用calculator應該計到
x = 5/2 or 4/25

Check solution:
(2(5/2)-5)^2 = (9(5/2)-3)(6(5/2)-15)
0 = 0
(2(4/25)-5)^2 = (9(4/25)-3)(6(5/2)-15)
13689/625 =/= 0

所以 x = 5/2