Answer
$\color{blue}{\left\{1-\sqrt3, 1+\sqrt3\right\}}$
Work Step by Step
Add $2$ to both sides:
$x-2x=2$
Add the square of one-half of the coefficient of $x$, which is $(\frac{-2}{2})^2=(-1)^2=1$, to obtain:
$x^2-2x+1=2+1
\\(x-1)^2=3$
Take the square root of both sides to obtain:
$\sqrt{(x-1)^2}=\pm \sqrt{3}
\\x-1 = \pm \sqrt3$
Add $1$ to both sides:
$x=1 \pm \sqrt3$
Thus, the solution set is: $\color{blue}{\left\{1-\sqrt3, 1+\sqrt3\right\}}$.
