How to do it ? Please help me~~ (Maths)

10x^2+9x-22=0 =
x^2 + 9/10 x - 22/10 = 0
x^2 +2 (x) (9 /20) + (9/20)^2 - (9/20)^2 - 22/10 = 0
(x + 9/20)^2 = (9/20)^2 + 22/10
(x + 9/20) ^2 = 81/400 + 22/10
(x + 9/20)^2 = 81/400 + 880/400
(x + 9/20)^2 = 961/400
(x + 9/20) = ± √ (961/400)
(x + 9/20) = ± (31/20)
x = ± (31/20) - 9/20
x = ± (22/20)
x = ± 11/10

you can use the quadratic formula x=[-b+/-√(b^2-4ac)]/2a

now, a=10, b=9, c=-22
so,

x={-9+/-√[(9^2-4(10)(-22)]} / 2(10)
=[-9+/-√(81+880)] /20
=[-9+/-√(961)] /20
=(-9+/-31)/20
=(-9+31)/20 or =(-9-31)/20
=11/10 =2

2006-10-18 18:08:44 補充：
aiya....痴埋左tim...咁樣寫清楚d=(-9 31)/20 =11/10 or =(-9-31)/20=2

2006-10-18 18:10:17 補充：
="=........漏左個負tim.呢個岩啦!!!!!=(-9 31)/20 =11/10 or =(-9-31)/20=-2

2006-10-24 19:44:14 補充：
the answer is wrong!!!wrongwrongwrong!!!!
it should be....x=11/10 or x=-2
如果x=-11/10...咁就唔等啦...
10x^2+9x-22=-19.8
而唔係等於0...
yahoo知識有個唔好ge地方就係...choose左出黎ge answer未必係岩ge!!!