#9x^2-5x-7# is of the form #ax^2+bx+c#, with #a=9#, #b=-5# and #c=-7#.
This has discriminant #Delta# given by the formula:
#Delta = b^2-4ac = (-5)^2-(4xx9xx-7) = 25+252#
#= 277#
Since #Delta > 0#, the quadratic equation has two distinct Real roots. Since #Delta# is not a perfect square (#277# is prime), those roots are irrational.
The roots are given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#=(-b+-sqrt(Delta))/(2a)#
#=(5+-sqrt(277))/18#
Notice the discriminant #Delta# is the expression under the square root.
So if #Delta < 0# the square root is not Real and the quadratic has no Real roots - It has a conjugate pair of distinct complex roots.
If #Delta = 0# then there is one repeated Real root.
If #Delta > 0# (as in our example), there are two distinct Real roots. If in addition #Delta# is a perfect square, then those roots are rational.
