Answer
Zeros: $\dfrac{-1+\sqrt{7}}{6}, \dfrac{-1-\sqrt{7}}{6}$
$x$-intercepts: $\dfrac{-1+\sqrt{7}}{6}, \dfrac{-1-\sqrt{7}}{6}$
Work Step by Step
To find the zeros of a function $g$, solve the equation $g(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $F(x)=0$:
$$3x^2+x-\dfrac{1}{2}=0$$
Taking $3$ as a common factor:
$$3\left(x^2+\dfrac{1}{3}x-\dfrac{1}{6}\right)=0$$
$$x^2+\dfrac{1}{3}x-\dfrac{1}{6}=0$$
Rearranging the Equation:
$$x^2+\dfrac{1}{3}x=\dfrac{1}{6}$$
The coefficient of $x^2$ is 1 and that of $x$ is$ \dfrac{1}{3}$, complete the square by adding $\left(\dfrac{1 \times \dfrac{1}{3}}{2}\right)^2 = \dfrac{1}{36}$
$$\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)=\dfrac{1}{6}+\dfrac{1}{36}$$
$$\left(x+\dfrac{1}{6}\right)^2 = \dfrac{7}{36}$$
$$\left(x+\dfrac{1}{6}\right)=\pm \dfrac{\sqrt{7}}{6}$$
$\therefore x+\dfrac{1}{6} = \dfrac{\sqrt{7}}{6} \hspace{20pt}\text{ or } \hspace{20pt} x+\dfrac{1}{6} = -\dfrac{\sqrt{7}}{6} $
Solve each equation to obtain the zeros of the function (which are the also the function's $x$-intercepts):
$x+\dfrac{1}{6} = \dfrac{\sqrt{7}}{6} \to x=\dfrac{-1+\sqrt{7}}{6} $
$x+\dfrac{1}{6} = -\dfrac{\sqrt{7}}{6} \to x= \dfrac{-1-\sqrt{7}}{6}$
