The problem asks us to find the quadratic equation whose roots are $\frac{1}{2}$ and $-\frac{1}{3}$.

1. Problem Description

The problem asks us to find the quadratic equation whose roots are $\frac{1}{2}$ and $-\frac{1}{3}$.

2. Solution Steps

We know that a quadratic equation can be written in the form $a(x - r_1)(x - r_2) = 0$, where $r_1$ and $r_2$ are the roots of the equation.

In this case, $r_1 = \frac{1}{2}$ and $r_2 = -\frac{1}{3}$.

So, the equation is $a(x - \frac{1}{2})(x + \frac{1}{3}) = 0$.

Expanding the expression, we get:

$a(x^2 + \frac{1}{3}x - \frac{1}{2}x - \frac{1}{6}) = 0$

$a(x^2 - \frac{1}{6}x - \frac{1}{6}) = 0$

To eliminate the fractions, we can choose $a = 6$:

$6(x^2 - \frac{1}{6}x - \frac{1}{6}) = 0$

$6x^2 - x - 1 = 0$

Therefore, the quadratic equation is $6x^2 - x - 1 = 0$.

3. Final Answer

D. $6x^2 - x - 1 = 0$