## Precalculus (6th Edition)

The solutions are $x=-\dfrac{1}{4}\pm\dfrac{\sqrt{97}}{4}$
$\dfrac{1}{2}x^{2}+\dfrac{1}{4}x-3=0$ Multiply the whole equation by $4$: $4\Big(\dfrac{1}{2}x^{2}+\dfrac{1}{4}x-3=0\Big)$ $2x^{2}+x-12=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=2$, $b=1$ and $c=-12$ Substitute the known values into the formula and evaluate: $x=\dfrac{-1\pm\sqrt{1^{2}-4(2)(-12)}}{2(2)}=\dfrac{-1\pm\sqrt{1+96}}{4}=...$ $...=\dfrac{-1\pm\sqrt{97}}{4}=-\dfrac{1}{4}\pm\dfrac{\sqrt{97}}{4}$ The solutions are $x=-\dfrac{1}{4}\pm\dfrac{\sqrt{97}}{4}$