#### Answer

$x=1\pm\sqrt{3}$

#### Work Step by Step

$\dfrac{1}{2}x^{2}-x-1=0$
Multiply the whole equation by $2$ to avoid using fractions in the quadratic formula. This will make the job easier.
$2\Big(\dfrac{1}{2}x^{2}-x-1=0\Big)$
$x^{2}-2x-2=0$
Now, use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=-2$ and $c=-2$.
Substitute:
$x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\dfrac{2\pm\sqrt{4+8}}{2}=...$
$...=\dfrac{2\pm\sqrt{12}}{2}=\dfrac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$