Answer
{$\frac{1}{6}\pm\frac{47}{6}i$}
Work Step by Step
Step 1: Comparing $3x^{2}-x+4=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=3$, $b=-1$ and $c=4$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a,b and c in the formula:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(3)(4)}}{2(3)}$
Step 4: $x=\frac{1 \pm \sqrt {1-48}}{6}$
Step 5: $x=\frac{1 \pm \sqrt {-47}}{6}$
Step 6: $x=\frac{1 \pm \sqrt {-1\times47}}{6}$
Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {47})}{6}$
Step 8: $x=\frac{1 \pm (i\times \sqrt {47})}{6}$
Step 9: $x=\frac{1 \pm i\sqrt {47}}{6}$
Step 10: $x=\frac{1}{6}\pm\frac{47i}{6}$
Step 11: Therefore, the solution set is {$\frac{1}{6}\pm\frac{47}{6}i$}.
