This equation's roots are complex numbers; it has no roots in the set of real numbers. To show this, compute the coordinates of the parabola's vertex. The x-coordinate is located at x = -b/2a (where a=5 and b=-4). To find the y-coordinate, plug the value of the x coordinate of the vertex into your quadratic equation.

Since the coefficient of the x^{2} term is positive (5), the parabola opens upward. If the y-coordinate of the vertex is positive, then the parabola will not cross the x-axis and will have no real roots. (The roots are the points where the graph crosses the x-axis)

To find the complex roots, use the quadratic formula:

x = (-b/2a) ± (1/2a)√(b^{2}-4ac)

For this problem, a = 5, b = -4, c = 1. Plug those values into the quadratic formula to get the answer.

Given that the roots are complex, the expression is not easily factored and the graph will not show complex roots. It may be that the teacher wants to the student to show that there are no real roots since the vertex is above the x-axis and the parabola opens upward. I'll amend the answer to include this possibility. Thanks, Robert.

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