Answer
{$\frac{1 - i\sqrt {7}}{4},\frac{1 + i\sqrt {7}}{4}$}
Work Step by Step
Step 1: Comparing $2x^{2}-x+1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=2$, $b=-1$ and $c=1$
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:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(2)(1)}}{2(2)}$
Step 4: $x=\frac{1 \pm \sqrt {1-8}}{4}$
Step 5: $x=\frac{1 \pm \sqrt {-7}}{4}$
Step 6: $x=\frac{1 \pm \sqrt {-1\times7}}{4}$
Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {7})}{4}$
Step 8: $x=\frac{1 \pm (i\times \sqrt {7})}{4}$
Step 9: $x=\frac{1 \pm i\sqrt {7}}{4}$
Step 10: $x=\frac{1 - i\sqrt {7}}{4}$ or $x=\frac{1 + i\sqrt {7}}{4}$
Step 11: Therefore, the solution set is {$\frac{1 - i\sqrt {7}}{4},\frac{1 + i\sqrt {7}}{4}$}.
