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