Question 702261
If x and y are roots of {{{ax^2+bx+c=0}}}, where x > y


then by the quadratic formula


{{{x = (-b+sqrt(b^2-4ac))/(2a)}}}


{{{y = (-b-sqrt(b^2-4ac))/(2a)}}}


So when you subtract y from x, you get


{{{x-y = (-b+sqrt(b^2-4ac))/(2a) - (-b-sqrt(b^2-4ac))/(2a)}}}


{{{x-y = (-b+sqrt(b^2-4ac) - (-b-sqrt(b^2-4ac)))/(2a)}}}


{{{x-y = (-b+sqrt(b^2-4ac) + b + sqrt(b^2-4ac))/(2a)}}}


{{{x-y = (2*sqrt(b^2-4ac))/(2a)}}}


{{{x-y = (sqrt(b^2-4ac))/a}}}