#### Answer

Option $(C)$
$x=\left\{-4, 3\right\}$

#### Work Step by Step

The equation in Option (C) has $a=1$ so it does not require Step 1 of the steps in solving using the completing the square method outlined in the book.
Add the square of one-half of the coefficient of $x$ which is$(\frac{1}{2})^2=\frac{1}{4}$, to both sides of the equation to obtain:
$x^2+x+\frac{1}{4}=12+\frac{1}{4}
\\(x+\frac{1}{2})^2=\frac{49}{4} + \frac{1}{4}
\\(x+\frac{1}{2})^2=\frac{49}{4}$
Take the square root of both sides to obtain:
$\sqrt{(x+\frac{1}{2})^2}=\pm \sqrt{\frac{49}{4}}
\\x+\frac{1}{2}=\pm \frac{7}{2}$
Subtract $\frac{1}{2}$ to both sides of the equation to obtain:
$x=-\frac{1}{2} \pm \frac{7}{2}
\\x_1=-\frac{1}{2}+\frac{7}{2} = \frac{6}{2}=3
\\x_2=-\frac{1}{2}-\frac{7}{2} = -\frac{8}{2}=-4$
Thus, the solutions are:
$\color{blue}{x=\left\{-4, 3\right\}}$