Answer
The solution set is $\left\{2\right\}$.
Work Step by Step
Add 4 on both sides of the equation to obtain:
$\sqrt{3(x+10)}=x+4$
Square both sides to obtain:
$3(x+10)=(x+4)^2
\\3(x)+3(10)=(x+4)^2
\\3x+30=(x+4)^2$
Use the rule $(a+b)^2 = a^2+2ab+b^2$ to obtain:
$3x+30=x^2+2(x)(4)+4^2
\\3x+30=x^2+8x+16$
Subtract $3x$ and $30$ on both sides of the equation, and then combine like terms to obtain:
$\begin{array}{ccc}
&3x+30-3x-30&= &x^2+8x+16-3x-30
\\&0 &= &x^2+5x-14
\end{array}$
Factor the trinomial to obtain:
$0=(x+7)(x-2)$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x+7=0 &\text{ or } &x-2=0
\\&x=-7 &\text{ or } &x=2
\end{array}$
$-7$ is an extraneous solution since it does not satisfy the original equation.
Thus, the solution set is $\left\{2\right\}$.
