Answer
{$\frac{-1 \pm \sqrt (97)}{4}$}
Work Step by Step
Step 1: Comparing $\frac{1}{2}x^{2}+\frac{1}{4}x-3=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=\frac{1}{2}$, $b=\frac{1}{4}$ and $c=-3$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt (b^{2}-4ac)}{2a}$
Step 3: Substituting the values of a,b and c in the formula:
$x=\frac{-(\frac{1}{4}) \pm \sqrt ((\frac{1}{4})^{2}-4(\frac{1}{2})(-3))}{2(\frac{1}{2})}$
Step 4: $x=\frac{-(\frac{1}{4}) \pm \sqrt (\frac{1}{16}+6)}{1}$
Step 5: $x=\frac{-(\frac{1}{4}) \pm \sqrt (\frac{97}{16})}{1}$
Step 6: $x={-(\frac{1}{4}) \pm \sqrt (97\times\frac{1}{16})}$
Step 7: $x={-(\frac{1}{4}) \pm (\sqrt (97)\times\sqrt (\frac{1}{16}))}$
Step 8: $x={-(\frac{1}{4}) \pm \frac{1}{4}\sqrt (97)}$
Step 9: $x={\frac{1}{4}(-1 \pm \sqrt (97)})$
Step 10: $x=\frac{-1 \pm \sqrt (97)}{4}$
Step 11: Therefore, the solution set is {$\frac{-1 \pm \sqrt (97)}{4}$}.
