Answer
The solution set is {$\frac{-1-\sqrt 3}{2},\frac{-1+\sqrt 3}{2}$}.
Work Step by Step
$2x^2+2x-1=0$
$a=2, b=2, c=1$
$D=b^2-4ac=(2)^2-4\cdot2(-1)=4+8=12>0$ The equation has two real solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-2\pm\sqrt 12}{2\cdot2}$ Use the quadratic formula.
$x=\frac{-2\pm2\sqrt 3}{4}$
$x=\frac{-1\pm\sqrt 3}{2}$
The solution set is {$\frac{-1-\sqrt 3}{2},\frac{-1+\sqrt 3}{2}$}.
