Answer
Zeros: $\dfrac{\sqrt{3}-1}{2} , \dfrac{-\sqrt{3}-1}{2}$
$x$-intercepts: $\dfrac{\sqrt{3}-1}{2} , \dfrac{-\sqrt{3}-1}{2}$
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 $f(x)=0$:
$$2x^2-1+2x=0$$
Rearranging the equation:
$$2x^2+2x-1=0$$
Comparing $2x^2+2x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 2, b=2 , c =-1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (2)^2-4 \times 2 \times -1 = 12$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-2 \pm \sqrt{12}}{2\times 2}$$
$$x=\dfrac{-2 \pm 2\sqrt{3}}{4}$$
$\therefore x =\dfrac{\sqrt{3}-1}{2}\hspace{20pt} \text{or} \hspace{20pt} x=\dfrac{-\sqrt{3}-1}{2}$
