方程有两根
Δ=(2m-3)^2-4m(m-2)≥0
-12m+9+8m≥0
m≤9/4
tan(α+β)
=(tana+tanb)/(1-tanatanb)
=-(2m-3)/m / (1-(m-2)/m)
=3-2m / (m-m+2)
=(3-2m)/2
≥(3-2*9/4)/2
=-3/4为什么（tana+tanb）=-(2m-3)/mtanatanb=(m-2)/m？这个根与系数关系没有学过么0.0ax^2+bx+c=0的两根是x1,x2则x1+x2 = -b/ax1x2 = c/a