Answer
Zeros: $-1 +\dfrac{\sqrt{10}}{2},-1 -\dfrac{\sqrt{10}}{2}$
$x$-intercepts: $-1 +\dfrac{\sqrt{10}}{2},-1 -\dfrac{\sqrt{10}}{2}$
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 $G(x)=0$:
$$2x(x+2)-3=0$$
Expanding the equation:
$$2x^2+4x-3=0$$
Comparing $2x^2+4x-3=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 2, b=4 , c =-3$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (4)^2-4 \times 2 \times -3 = 40$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-4\pm \sqrt{40}}{2\times 2}$$
$$x=\dfrac{-4 \pm 2\sqrt{10}}{4}$$
$\therefore x =-1 +\dfrac{\sqrt{10}}{2}\hspace{20pt} \text{or} \hspace{20pt} x=-1 -\dfrac{\sqrt{10}}{2}$