## Algebra: A Combined Approach (4th Edition)

$x=1\pm\sqrt{3}$
$\dfrac{1}{2}x^{2}-x-1=0$ 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: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\dfrac{2\pm\sqrt{4+8}}{2}=...$ $...=\dfrac{2\pm\sqrt{12}}{2}=\dfrac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$