Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $x^{2} + x + \frac{1}{4} = 0$.

(D) For the quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Comparing $x^{2} + x + \frac{1}{4} = 0$ with the standard form,we have $a = 1$,$b = 1$,and $c = \frac{1}{4}$.
Substituting these values into the formula: $D = (1)^{2} - 4(1)(\frac{1}{4}) = 1 - 1 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real,rational,and equal.