#x^2-3x-5# is of the form #ax^2+bx+c#, with #a=1#, #b=-3# and #c=-5#.
Its discriminant #Delta# is given by the formula:
#Delta = b^2-4ac = (-3)^2 - (4xx1xx-5) = 9+20 = 29#
Since #Delta > 0# the equation has two distinct real roots, but since #29# is not a perfect square, those roots are irrational. That is they are not expressible as #p/q# for some integers #p# and #q#.
The solutions of #x^2-3x-5 = 0# are given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)#
#=(3+-sqrt(29))/2#
Notice that the discriminant #Delta# is the part under the square root. Hence if #Delta > 0# we get the two distinct real roots. If #Delta = 0# we get one repeated real root. If #Delta < 0# then the equation has no real roots (it has two distinct complex roots).
