Answer
$\color{blue}{\left\{3, 4\right\}}$
Work Step by Step
Subtract $12$ to both sides:
$x^2-7x=-12$
Add the square of one-half of the coefficient of $x$, which is $(\frac{-7}{2})^2=\frac{49}{4}$, to obtain:
$x^2-7x+\frac{49}{4}=-12+\frac{49}{4}
\\(x-\frac{7}{2})^2=-\frac{48}{4} + \frac{49}{4}
\\(x-\frac{7}{2})^2=\frac{1}{4}$
Take the square root of both sides to obtain:
$\sqrt{(x-\frac{7}{2})^2}=\pm \sqrt{\frac{1}{4}}
\\x-\frac{7}{2} = \pm \frac{1}{2}$
Add $\frac{7}{2}$ to both sides:
$x=\frac{7}{2} \pm \frac{1}{2}
\\x_1=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3
\\x_2=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4$
Thus, the solution set is: $\color{blue}{\left\{3, 4\right\}}$.
