Answer: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
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