Answer
The $x$-intercepts of this equation are $x=-\dfrac{3}{4}\pm\dfrac{\sqrt{41}}{4}$
Work Step by Step
$y=\dfrac{x^{2}}{3}+\dfrac{x}{2}-\dfrac{2}{3}$
To find the $x$-intercepts of this equation, set $y$ equal to $0$ and solve for $x$:
$\dfrac{x^{2}}{3}+\dfrac{x}{2}-\dfrac{2}{3}=0$
Multiply the whole equation by $6$:
$6\Big(\dfrac{x^{2}}{3}+\dfrac{x}{2}-\dfrac{2}{3}=0\Big)$
$2x^{2}+3x-4=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=2$, $b=3$ and $c=-4$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(2)(-4)}}{2(2)}=\dfrac{-3\pm\sqrt{9+32}}{4}=...$
$...=\dfrac{-3\pm\sqrt{41}}{4}=-\dfrac{3}{4}\pm\dfrac{\sqrt{41}}{4}$
The $x$-intercepts of this equation are $x=-\dfrac{3}{4}\pm\dfrac{\sqrt{41}}{4}$
