为什么一元二次方程中x1与x2相减的绝对值等于根号下△除以a的绝对值?

|x1-x2|=√(x1-x2)^2=√(x1^2-2x1x2+x2^2)=√(x1^2+2x1x2+x2^2-4x1x2)
=√[(x1+x2)^2-4x1x2]
=√[(-b/a)^2-4c/a]
=√[(b^2-4ac)/a^2]=√△/|a|