#### Answer

$\color{blue}{\left\{-\frac{7}{2}, 4\right\}}$

#### Work Step by Step

Divide $2$ to both sides:
$x^2-\frac{1}{2}x-14=0$
Add $14$ to both sides:
$x-\frac{1}{2}x=14$
Add the square of one-half of the coefficient of $x$, which is $(\dfrac{-\frac{1}{2}}{2})^2=(\frac{-1}{4})^2=\frac{1}{16}$, to obtain:
$x^2-\frac{1}{4}x+\frac{1}{16}=14+\frac{1}{16}
\\(x-\frac{1}{4})^2=\frac{224}{16} + \frac{1}{16}
\\(x-\frac{1}{4})^2=\frac{225}{16}$
Take the square root of both sides to obtain:
$\sqrt{(x-\frac{1}{4})^2}=\pm \sqrt{\frac{225}{16}}
\\x-\frac{1}{4} = \pm \frac{15}{4}$
Add $\frac{1}{4}$ to both sides:
$x=\frac{1}{4} \pm \frac{15}{4}
\\x_1=\frac{1}{4}-\frac{15}{4}=\frac{-14}{4}=-\frac{7}{2}
\\x_2=\frac{1}{4}+\frac{15}{4}=\frac{16}{4}=4$
Thus, the solution set is: $\color{blue}{\left\{-\frac{7}{2}, 4\right\}}$.