Answer
The solution set is {$\frac{1-{i}\sqrt 31}{8},\frac{1+{i}\sqrt 31}{8}$}.
No real solution.
Work Step by Step
$4y^2-y+2=0$
$a=4, b=-1, c=2$
$D=b^2-4ac=(-1)^2-4\cdot4(2)=1-32=-31<0$ The equation has two complex solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{1\pm\sqrt -31}{2\cdot4}$ Use the quadratic formula.
$x=\frac{1\pm{i}\sqrt 31}{8}$
$x=\frac{1+{i}\sqrt 31}{8}$,$x=\frac{1-{i}\sqrt 31}{8}$
The solution set is {$\frac{1-{i}\sqrt 31}{8},\frac{1+{i}\sqrt 31}{8}$}. No real solution.
