Answer
Zeros: $2\sqrt{2}-\dfrac{3}{2}, -2\sqrt{2}-\dfrac{3}{2}$
$x$-intercepts: $2\sqrt{2}-\dfrac{3}{2}, -2\sqrt{2}-\dfrac{3}{2}$
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 $F(x)=0$:
$$(2x+3)^2-32=0$$
Using the Square Root Method
$$(2x+3)^2=32$$
$$ (2x+3) = \pm \sqrt{32}$$
$\sqrt{32} \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$
$$(2x+3) = \pm 4 \sqrt{2}$$
$\therefore 2x+3 = 4 \sqrt{2} \text{ or } 2x+3 = -4 \sqrt{2} $
$2x+3 = 4 \sqrt{2} \to 2x = 4 \sqrt{2}-3 \to x = 2\sqrt{2} - \dfrac{3}{2}$
$2x+3 = -4 \sqrt{2} \to 2x =-4 \sqrt{2}-3 \to x = -2 \sqrt{2}-\dfrac{3}{2}$
