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).
