add math

For the equation
2(2x+3)^2 - 3 l2x+3l - 20 = 0

Put y = l2x+3l , than the given equation becomes
2y^2 - 3y - 20 = 0
(2y + 5) (y - 4) = 0
2y + 5 = 0 or y - 4 = 0
y = -5/2 or y = 4

Hence
l2x+3l = -5/2 (rejected since l2x+3l >0) or l2x+3l = 4

So 2x + 3 = 4 or 2x + 3 = -4
2x = 1 or 2x = -7
x = 0.5 or x = -3.5

2006-10-14 00:48:48 補充：
This is the fastest methodand I got A grade in both Maths, A Maths and also Pure Maths

if x>-3/2
then
2(2x+3)^2 - 3 l2x+3l - 20 = 0
2(2x+3)^2 - 3 (2x+3) - 20 = 0

let y=2x+3
2y^2-3y-20=0
(2y+5)(y-4)=0
y=-5/2 or y=4
2x+3=-5/2 or 2x+3=4
x=-11/4 (rejected) or x=1/2

when x=-3/2
2(2x+3)^2 - 3 l2x+3l - 20 = 0
-20=0
contradiction

when x<-3/2
2(2x+3)^2 - 3 l2x+3l - 20 = 0
2(2x+3)^2 +3 (2x+3) - 20 = 0
let y=2x+3
2y^2+3y-20=0
(2y-5)(y+4)=0
y=5/2 or y=-4
2x+3=5/2 or 2x+3=-4
x=-1/4 (rejected) or x=-7/2
so x=1/2 or x=-7/2

因為 |2x+3|^2=(2x+3)^2
所以姑且當　y=|2x+3|

Solve 2y^2 - 3y -20 =0
得到 y 之後（有兩個數值）再用番條式
y= |2x+3|

Solve 番個 x 出黎
應該會得到有四個答案。

（懂得呢個方法，好多問題都迎刃而解）