Answer
$\left\{-2\pm\sqrt{19}\right\}$
Work Step by Step
Completing the square of the left-hand expression of $x^2+bx=c$, the given equation, $
x^2+4x=15
,$ is equivalent to
\begin{align*}
x^2+4x+\left(\dfrac{4}{2}\right)^2&=15+\left(\dfrac{4}{2}\right)^2
&(\text{add }\left(\dfrac{b}{2}\right)^2)
\\\\
x^2+4x+4&=15+4
\\
(x+2)^2&=19
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}
x+2&=\pm\sqrt{19}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
x&=-2\pm\sqrt{19}
\end{align*}
Hence, the solution set of $
x^2+4x=15
$ is $
\left\{-2\pm\sqrt{19}\right\}
$.
