Answer:
This function would have two distinct real roots.
Step-by-step explanation:
Consider a quadratic function [tex]f(x) = a\, x^{2} + b\, x + c[/tex].
The determinant of this function would be [tex]\Delta = b^{2} - 4\, a\, c[/tex].
The (only) two roots of this quadratic function would be:
[tex]\begin{aligned} x_{1} = \frac{-b - \sqrt{\Delta}}{2\, a}\end{aligned}[/tex], and
[tex]\begin{aligned} x_{2} = \frac{-b + \sqrt{\Delta}}{2\, a}\end{aligned}[/tex].
Because of the square root [tex]\sqrt{\Delta}[/tex] in the two expressions, [tex]x_{1}[/tex] and [tex]x_{2}[/tex] would take real values if and only if [tex]\Delta \ge 0[/tex] (determinant is nonnegative.) If [tex]\Delta < 0[/tex], this quadratic function would not have any real root.
Since the only difference between the two roots [tex]x_{1}[/tex] and [tex]x_{2}[/tex] is [tex](1/a)\, \sqrt{\Delta}[/tex], these two roots would repeat one another if [tex]\Delta = 0[/tex] (determinant is zero.)
Otherwise, if [tex]\Delta > 0[/tex], this quadratic function would have two distinct real roots.
