Answer
The solution set is {$\frac{-1-{i}\sqrt 15}{8},\frac{-1+{i}\sqrt 15}{8}$}. No real solution.
Work Step by Step
$4t^2+t+1=0$
$a=4, b=1, c=1$
$D=b^2-4ac=(1)^2-4\cdot4(1)=1-16=-15<0$ The equation has two complex solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-1\pm\sqrt -15}{2\cdot4}$ Use the quadratic formula.
$x=\frac{-1\pm{i}\sqrt 15}{8}$
$x=\frac{-1+{i}\sqrt 15}{8}$,$x=\frac{-1-{i}\sqrt 15}{8}$
The solution set is {$\frac{-1-{i}\sqrt 15}{8},\frac{-1+{i}\sqrt 15}{8}$}. No real solution.
