试题分析:解一元二次方程的方法有:1.配方法;2.公式法;3.十字相乘法;方法一(求根公式):去括号,将方程化为2x
2-x+6=0,x=
, x
1=2,x
2=
;方法二(因式分解):x(2x+3)="2(2x+3)," x(2x+3)-2(2x+3)="0,(x-2)(2x+3)=0," x
1=2,x
2=
.
试题解析:方法一(求根公式):
去括号,将方程化为2x
2-x-6="0,"
b
2-4ac=(-1)
2-4×2×(-6)=49,
x=
,
x
1=2,x
2=
.
方法二(因式分解):
x(2x+3)="2(2x+3),"
x(2x+3)-2(2x+3)=0,
(x-2)(2x+3)="0,"
x
1=2,x
2=
.