f(x) = 2x^2+5x+5 is of the form ax^2+bx+c with a=2, b=5 and c=5.
This has discriminant Delta given by the formula:
Delta = b^2-4ac = 5^2 - (4xx2xx5) = 25 - 40 = -15
Since the discriminant is negative, f(x) = 0 has no real roots. It only has complex ones.
The quadratic formula still works, giving the roots as:
x = (-b+-sqrt(Delta))/(2a) = (-5+-sqrt(-15))/(2*2)
=(-5+-i sqrt(15))/4
In general the various cases for the different values of the discriminant are as follows:
Delta > 0 The quadratic equation has two distinct real roots. If Delta is a perfect square (and the coefficients of the quadratic are rational) then the roots are rational too.
Delta = 0 The quadratic equation has one repeated real root. It is a perfect square trinomial.
Delta < 0 The quadratic equation has no real roots. It has a conjugate pair of distinct complex roots.
