Answer
{$\frac{1-\sqrt {33}}{4},\frac{1+\sqrt {33}}{4}$}
Work Step by Step
Step 1: Comparing $2y^{2}-y-4=0$ to the standard form of a quadratic equation, $ay^{2}+by+c=0$, we find:
$a=2$, $b=-1$ and $c=-4$
Step 2: The quadratic formula is:
$y=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b, and c in the formula:
$y=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(2)(-4)}}{2(2)}$
Step 4: $y=\frac{1 \pm \sqrt {1+32}}{4}$
Step 5: $y=\frac{1 \pm \sqrt {33}}{4}$
Step 6: $y=\frac{1-\sqrt {33}}{4}$ or $y=\frac{1+\sqrt {33}}{4}$
Step 7: Therefore, the solution set is {$\frac{1-\sqrt {33}}{4},\frac{1+\sqrt {33}}{4}$}.
