Answer
$\{\frac{1+i\sqrt 23}{6},\frac{1-i\sqrt 23}{6}\}$
Work Step by Step
$3x^{2}-x+2=0$
Using quadratic formula,
$x= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=3, b=-1, c=2$
$x= \frac{-(-1)±\sqrt ((-1)^{2}-4(3)(2))}{2(3)}$
$x= \frac{1±\sqrt (1-24)}{6}$
$x= \frac{1±\sqrt (-23)}{6}$
Using $\sqrt -b = i\sqrt b$
$x= \frac{1±i\sqrt 23}{6}$
$x= \frac{1+i\sqrt 23}{6}$ or $x= \frac{1-i\sqrt 23}{6}$
