Answer
$\{-5,-1,3\}$
Work Step by Step
We have to solve the quadratic equation:
$$(x^2+2x)^2-14(x^2+2x)=15.\tag1$$
We substitute $y=x^2+2x$ and bring the equation to the standard form:
$$y^2-14y-15=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y+1)(y-15)&=0\\
y+1=0&\text{ or }y-15=0\\
y=-1&\text{ or }y=15.
\end{align*}$$
We determine $x$ using the values of $y$ that we found.
$\textbf{Case 1: }$ $y=-1$
$$\begin{align*}
x^2+2x&=-1\\
x^2+2x+1&=0\\
(x+1)^2&=0\\
x+1&=0\\
x&=-1;
\end{align*}$$
$\textbf{Case 2: }$ $y=15$
$$\begin{align*}
x^2+2x&=15\\
x^2+2x-15&=0\\
(x-3)(x+5)&=0\\
x-3=0&\text{ or }x+5=0\\
x=3&\text{ or }x=-5.
\end{align*}$$
Check the solutions:
$$\begin{align*}
x&=-1\\
((-1)^2+2(-1))^2-14((-1)^2+2(-1))&\stackrel{?}{=}15\\
1+15&\stackrel{?}{=}15\\
15&=15\checkmark\\\\
x&=3\\
(3^2+2(3)^2-14(3^2+2(3))&\stackrel{?}{=}15\\
225-210&\stackrel{?}{=}15\\
15&=15\checkmark\\\\
x&=-5\\
((-5)^2+2(-5))^2-14((-5)^2+2(-5))&\stackrel{?}{=}15\\
225-210&\stackrel{?}{=}15\\
15&=15\checkmark
\end{align*}$$
The solution set of the equation is:
$$\{-5,-1,3\}.$$
