Answer
Zeros: $-1 +\dfrac{2\sqrt{3}}{3},-1 -\dfrac{2\sqrt{3}}{3}$
$x$-intercepts: $-1 +\dfrac{2\sqrt{3}}{3},-1 -\dfrac{2\sqrt{3}}{3}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $F(x)=0$:
$$3x(x+2)-1=0$$
Expanding the equation:
$$3x^2+6x-1=0$$
Comparing $3x^2+6x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 3, b=6 , c =-1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (6)^2-4 \times 3 \times -1 = 48$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-6\pm \sqrt{48}}{2\times 3}$$
$$x=\dfrac{-6\pm 4\sqrt{3}}{6}$$
$\therefore x =-1 +\dfrac{2\sqrt{3}}{3}\hspace{20pt} \text{or} \hspace{20pt} x=-1 -\dfrac{2\sqrt{3}}{3}$
