Answer
$\left\{-1\pm\sqrt{5}\right\}$
Work Step by Step
Completing the square of the left-hand expression by adding $\left(\dfrac{b}{2}\right)^2$, the given equation, $
x^2+2x=4
,$ is equivalent to
\begin{align*}
x^2+2x+\left(\dfrac{2}{2}\right)^2&=4+\left(\dfrac{2}{2}\right)^2
\\\\
x^2+2x+1&=4+1
\\
(x+1)^2&=5
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}
x+1&=\pm\sqrt{5}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
x&=-1\pm\sqrt{5}
.\end{align*}
Hence, the solution set of $
x^2+2x=4
$ is $
\left\{-1\pm\sqrt{5}\right\}
$.
