Answer
$x = \frac{-3±\sqrt{41}}{4}$
Work Step by Step
$\frac{x^2}{3} + \frac{x}{2} = \frac{2}{3}$
Multiply the equation by the least common multiple $6$.
$\frac{x^2}{3}⋅6 + \frac{x}{2}⋅6 = \frac{2}{3}⋅6$
$2x^2+3x = 4$
$2x^2+3x-4 = 0$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=2$, $b=3$, $c=-4$
$x = \frac{-3±\sqrt{3^2-(4⋅2⋅-4)}}{(2⋅2)}$
$x = \frac{-3±\sqrt{9-(-32)}}{4}$
$x = \frac{-3±\sqrt{41}}{4}$
