以知方程

2x^2-5x+m=0
由韦达定理
x1+x2=-(-5/2)=5/2
x1*x2=m/2

(x1-x2)^2=x^1+x^2-2x1x2=(x1+x2)^2-4x1x2=25/4-m^2=(25-4m^2)/4
x1-x2=±√(25-4m^2)/2

x1^2-x2^2=(x1+x2)(x1-x2)=±m√(25-4m^2)/4

2.5
m/2

±（根号下25-8m）/2

±（5*根号下25-8m）/4

5/2
m/2
±（根号下25-8m）/2
±（5*根号下25-8m）/4