Answer
Zeros: $\dfrac{1}{2}, -\dfrac{2}{3}$
$x$-intercepts: $\dfrac{1}{2}, -\dfrac{2}{3}$
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 $H(x)=0$:
$$6x^2+x-2=0$$
Comparing $6x^2+x-2=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 6, b=1 , c =-2$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (1)^2-4 \times 6 \times -2 = 49$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-1\pm \sqrt{49}}{2\times 6}$$
$$x=\dfrac{-1\pm 7}{12}$$
Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{1}{2}$ and $-\dfrac{2}{3}$.
