Answer
$\displaystyle \{\frac{-1-\sqrt{7}}{6},\frac{-1+\sqrt{7}}{6}\}$
Work Step by Step
$3x^{2}+x-\displaystyle \frac{1}{2}=0$
If we multiply the equation with 12, the first term becomes a square, and we get rid of the fraction....
$36x^{2}+12x-6=0$
The first two terms can be recognized as
$(6x)^{2}+2\cdot(6x)\cdot(1)$
The third term missing for a perfect square is $(1)^{2}$, so we rewrite the equation
$(6x)^{2}+2\cdot(6x)\cdot(1)+1^{2}\quad-1-6=0$
$(6x+1)^{2}-7=0$
$(6x+1)^{2}=7$
.. it follows that
$6x+1=\pm\sqrt{7}$
$6x=-1\pm\sqrt{7}$
$x= \displaystyle \frac{-1\pm\sqrt{7}}{6}$
Solution set = $\displaystyle \{\frac{-1-\sqrt{7}}{6},\frac{-1+\sqrt{7}}{6}\}$
