Answer
$$\left\{-\dfrac{1}{2} - \dfrac{\sqrt{5}}{2}, -\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right\}$$
Work Step by Step
Add $1$ to both sides to obtain:
$$x^2+x=1$$
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&=1+\left(\frac{1}{2}\right)^2\\\\
x^2+x+\frac{1}{4}&=1+\frac{1}{4}\\\\
\left(x+\frac{1}{2}\right)^2&=\frac{4}{4}+\frac{1}{4}\\\\
\left(x+\frac{1}{2}\right)^2&=\frac{5}{4}\\\\
\end{align*}
Take the square root of both sides:
\begin{align*}
\sqrt{\left(x+\frac{1}{2}\right)^2}&=\pm\sqrt{\frac{5}{4}}\\\\
x+\frac{1}{2}&=\pm\frac{\sqrt{5}}{2}\\\\
x&=-\frac{1}{2}\pm\frac{\sqrt{5}}{2}\\\\
\end{align*}
Thus, the solution set is:
$$\left\{-\dfrac{1}{2} - \dfrac{\sqrt{5}}{2}, -\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right\}$$
