Respuesta :

Recall that given the equation of the second degree (or quadratic)
 ax ^ 2 + bx + c
 Its solutions are:
 x = (- b +/- root (b ^ 2-4ac)) / 2a
 discriminating: 
 d = root (b ^ 2-4ac)
 If d> 0, then the two roots are real (the radicand of the formula is positive). 
 If d = 0, then the root of the formula is 0 and, therefore, there is only one solution that is real and of multiplicity 2 (it is a double root).
 If d <0, then the two roots are complex and, in addition, one is the conjugate of the other. That is, if one solution is x1 = a + bi, then the other solution is x2 = a-bi (we are assuming that a, b, c are real).
 One solution:
 A cut point with the x axis
 Two solutions:
 Two cutting points with the x axis.
 Complex solutions:
 Does not cut to the x axis