## Precalculus (6th Edition)

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