Answer
Zeros: $\dfrac{5 \sqrt{3}}{3}+\dfrac{2}{3}$
$x$-intercepts: $ -\dfrac{5 \sqrt{3}}{3}+\dfrac{2}{3}$
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 $g(x)=0$:
$$(3x-2)^2-75=0$$
Using the Square Root Method
$$(3x-2)^2=75$$
$$ (3x-2) = \pm \sqrt{75}$$
$\sqrt{75} =\sqrt{25 \times 3} = \sqrt{25} \sqrt{3} = 5 \sqrt{3}$
$$(3x-2) = \pm 5 \sqrt{3}$$
$\therefore 3x-2 = 5\sqrt{3} \text{ or } 3x-2 = -5\sqrt{3} $
$3x-2 = 5\sqrt{3} \to 3x = 5 \sqrt{3}+2 \to x = \dfrac{5 \sqrt{3}}{3}+\dfrac{2}{3}$
$3x-2 = -5\sqrt{3} \to 3x =-5 \sqrt{2}+2 \to x = -\dfrac{5 \sqrt{3}}{3}+\dfrac{2}{3}$
