Answer
The solution set is $\left\{ -1 \right\}$.
Work Step by Step
Adding 4 to both sides, we will get
$\begin{align}
& \sqrt{x+10}-4+4=x+4 \\
& \sqrt{x+10}=x+4
\end{align}$
The above equation can be squared using the following property:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Squaring both sides of the equation, we get:
$\begin{align}
& \sqrt{x+10}=x+4 \\
& {{\left( \sqrt{x+10} \right)}^{2}}={{\left( x+4 \right)}^{2}} \\
& x+10={{x}^{2}}+8x+16 \\
& {{x}^{2}}+7x+6=0
\end{align}$
Now, we will solve the quadratic equation
$\begin{align}
& {{x}^{2}}+7x+6=0 \\
& {{x}^{2}}+6x+x+6=0 \\
& x\left( x+6 \right)+1\left( x+6 \right)=0 \\
& \left( x+1 \right)\left( x+6 \right)=0
\end{align}$
This implies,
$\begin{matrix}
\left( x+1 \right)=0\text{ or }\left( x+6 \right)=0 \\
x=-1\text{ or }x=-6 \\
\end{matrix}$
Now, check the results
For $x=-1$
$\begin{align}
& \sqrt{x+10}-4=x \\
& \sqrt{-1+10}-4=-1 \\
& \sqrt{9}-4=-1 \\
& -1=-1
\end{align}$
It is true for $x=-1$
For $x=-6$
$\begin{matrix}
\sqrt{x+10}-4=x \\
\sqrt{-6+10}-4=-6 \\
\sqrt{4}-4=-6 \\
2-4=-6 \\
-2\ne -6 \\
\end{matrix}$
It is not true for $x=-6$.
Thus, the exact solution is $\left\{ -1 \right\}$.
