#### Answer

$\left\{2-\sqrt{3}, 2+\sqrt{3}\right\}$

#### Work Step by Step

Put all terms with a variable on the left side of the equation and the constant on the right side (remember that when a term is moved from one side to the other, it changes its sign) to obtain:
\begin{align*}
x^2-3x-x&=-1\\
x^2-4x&=-1
\end{align*}
RECALL:
To complete the square for $x^2 +mb$, just add $\left(\dfrac{m}{2}\right)^2$.
Use the rule above to complete the square (note that to retain the equality, whatever is added on the left side of the equation must also be added to the right side of the equation) and obtain:
\begin{align*}
x^2-4x+\left(\frac{-4}{2}\right)^2&=-1+\left(\frac{-4}{2}\right)^2\\\\
x^2-4x+\left(-2\right)^2&=-1+\left(-2\right)^2\\\\
x^2-4x+4&=-1+4\\\\
(x-2)^2&=3
\end{align*}
Take the square root of both sides to obtain:
\begin{align*}
\sqrt{(x-2)^2}&=\pm \sqrt{3}\\\\
x-2&=\pm \sqrt{3}\\\\
x&=2\pm \sqrt{3}\\\\
\end{align*}
Thus, the solutions are: $\left\{2-\sqrt{3}, 2+\sqrt{3}\right\}$