数学问题（加10分）

x^2+8x-20=0
(x+10)(x-2)=0
x1=-10,x2=2

(2y-5）^2-(2y-5)-72=0
[(2y-5)+8][(2y-5)-9]=0
2y-5=-8
y1=-3/2
2y-5=9
y2=7

1.(x+10)(x-2)=0 x=-10或x=2
2.设2y-5为z,
(2y-5）^2-(2y-5)-72=0为z^2-z-72=0
=>(z+8)(z-9)=0
=>(2y-5+8)(2y-5-9)=0
=>(2y+3)(y-7)=0 x=-3/2或x=7

1.(2y-5）^2-(2y-5)-72=0
[(2y-5)+8][(2y-5)-9]=0
2y-5=-8
y1=-3/2
2y-5=9
y2=7

1:
原式＝ x^2+8x+16+4
=(x+4)^2+4
=(x+4+2)(x+4-2)
=(x+6)(x-2)
2:
用十字相乘法：
原式=[（2y-5)-9][(2y-5)+8]
=(2y-14)(2y+3)

^这个符号是什么意思啊？