Answer
The solution of the equation is $ x\in \left( -\infty,-4 \right)\cup \left( 2,\infty \right)$.
Work Step by Step
Consider the equality for equation (1):
$\begin{align}
& {{x}^{2}}+2x+3=11 \\
& {{x}^{2}}+2x-8=0
\end{align}$
Now, we will solve the quadratic equation as below:
$\begin{align}
& {{x}^{2}}+2x-8=0 \\
& \left( x+4 \right)\left( x-2 \right)=0
\end{align}$
Therefore, $ x=-4,2$.
Hence, the intervals in which the inequality can be considered are $\left( -\infty,-4 \right),\text{ }\left( -4,2 \right),\text{ and }\left( 2,\infty \right)$.
Now, test the values from these intervals, such that only $\left( -\infty,-4 \right),\left( 2,\infty \right)$ stands true.
