Answer
$$\left\{\dfrac{1}{2} - \dfrac{\sqrt{21}}{2}, \dfrac{1}{2}+\dfrac{\sqrt{21}}{2}\right\}$$
Work Step by Step
Recall:
To complete the square of $x^2+bx$, add $\left(\dfrac{b}{2}\right)^2$.
The given equation has $b=-1$.
Complete the square by adding $\left(\dfrac{-1}{2}\right)^2$ to both sides of the equation to obtain:
\begin{align*}
x^2-x+\left(\frac{-1}{2}\right)^2&=5+\left(\frac{-1}{2}\right)^2\\\\
x^2-x+\frac{1}{4}&=5+\frac{9}{4}\\\\
\left(x-\frac{1}{2}\right)^2&=\frac{20}{4}+\frac{1}{4}\\\\
\left(x-\frac{1}{2}\right)^2&=\frac{21}{4}\\\\
\end{align*}
Take the square root of both sides:
\begin{align*}
\sqrt{\left(x-\frac{1}{2}\right)^2}&=\pm\sqrt{\frac{21}{4}}\\\\
x-\frac{1}{2}&=\pm\frac{\sqrt{21}}{2}\\\\
x&=\frac{1}{2}\pm\frac{\sqrt{21}}{2}\\\\
\end{align*}
Thus, the solution set is:
$$\left\{\dfrac{1}{2} - \dfrac{\sqrt{21}}{2}, \dfrac{1}{2}+\dfrac{\sqrt{21}}{2}\right\}$$
