#### Answer

{$\frac{1}{4}\pm\frac{\sqrt {31}}{4}i$}

#### Work Step by Step

We will use the quadratic formula to solve the equation over a set of complex numbers.
Step 1: Comparing $2x^{2}-x+4=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=2$, $b=-1$ 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:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(2)(4)}}{2(2)}$
Step 4: $x=\frac{1 \pm \sqrt {1-32}}{4}$
Step 5: $x=\frac{1 \pm \sqrt {-31}}{4}$
Step 6: $x=\frac{1 \pm \sqrt {-1\times31}}{4}$
Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {31})}{4}$
Step 8: As $i=\sqrt {-1}$, the equation becomes:
$x=\frac{1 \pm (i\times \sqrt {31})}{4}$
Step 9: $x=\frac{1 \pm i\sqrt {31}}{4}$
Step 10: $x=\frac{1}{4}\pm\frac{\sqrt {31}}{4}i$
Step 11: Therefore, the solution set is {$\frac{1}{4}\pm\frac{\sqrt {31}}{4}i$}.