Answer
{$\frac{1 - i\sqrt {31}}{8},\frac{1 + i\sqrt {31}}{8}$}
Work Step by Step
Step 1: Comparing $4x^{2}-x+2=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=4$, $b=-1$ and $c=2$
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:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(4)(2)}}{2(4)}$
Step 4: $x=\frac{1 \pm \sqrt {1-32}}{8}$
Step 5: $x=\frac{1 \pm \sqrt {-31}}{8}$
Step 6: $x=\frac{1 \pm \sqrt {-1\times31}}{8}$
Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {31})}{8}$
Step 8: $x=\frac{1 \pm (i\times \sqrt {31})}{8}$
Step 9: $x=\frac{1 \pm i\sqrt {31}}{8}$
Step 10: $x=\frac{1 - i\sqrt {31}}{8}$ or $x=\frac{1 + i\sqrt {31}}{8}$
Step 11: Therefore, the solution set is {$\frac{1 - i\sqrt {31}}{8},\frac{1 + i\sqrt {31}}{8}$}.
