#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.
