## College Algebra (10th Edition)

The solution set is {$\frac{1-{i}\sqrt 31}{8},\frac{1+{i}\sqrt 31}{8}$}. No real solution.
$4y^2-y+2=0$ $a=4, b=-1, c=2$ $D=b^2-4ac=(-1)^2-4\cdot4(2)=1-32=-31<0$ 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.