Answer
$$\text{H}$$
Work Step by Step
The $x$-intercepts of a parabola can be found by setting $y=0$ then solving for $x$:
\begin{align*}
y&=6x^2-4x-3\\
0&=6x^2-4x-3\\
\end{align*}
Recall:
The discriminant $b^2-4ac$ can tell the number of solutions/roots of a quadratic equation.
(1) If $b^2-4ac >0$, then the quadratic equation has two real roots.
(2) If $b^2-4ac<0$, then the quadratic equation has no real roots.
(3) If $b^2-4ac=0$, then the quadratic equation has one real solution.
The quadratic equation above has $a=6, b=-4,$ and $c=-3$.
Find the determinant to obtain:
\begin{align*}
b^2-4ac&= (-4)^2-4(6)(-3)\\
&=16+72\\
&=88
\end{align*}
The discriminant is greater than $0$ therefore the quadratic equation has $2$ real roots.
The roots of a quadratic equation are also its $x$-intercepts.
Thus, the given parabola has $2$ $x$-intercepts.
