Answer
{$-\frac{7}{2},4$}
Work Step by Step
Step 1: Comparing $2x^{2}-x-28=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=2$, $b=-1$ and $c=-28$
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:
$x=\frac{-(-1) \pm \sqrt ((-1)^{2}-4(2)(-28))}{2(2)}$
Step 4: $x=\frac{1 \pm \sqrt (1+224)}{4}$
Step 5: $x=\frac{1 \pm \sqrt (225)}{4}$
Step 6: $x=\frac{1 \pm15}{4}$
Step 7: $x=\frac{1+15}{4}$ or $x=\frac{1-15}{4}$
Step 8: $x=\frac{16}{4}$ or $x=\frac{-14}{4}$
Step 9: $x=4$ or $x=-\frac{7}{2}$
Step 10: Therefore, the solution set is {$-\frac{7}{2},4$}.
