Answer
$x=3$
Work Step by Step
Squaring both sides of the equation and then using the properties of equality, we obtain:
\begin{array}{l}\require{cancel}\left(
\sqrt{x^2+27}
\right)^2=\left(
x+3
\right)^2
\\\\
x^2+27=(x)^2+2(x)(3)+(3)^2
\\\\
x^2+27=x^2+6x+9
\\\\
(x^2-x^2)-6x+(27-9)=0
\\\\
-6x+18=0
\\\\
-6x=-18
\\\\
x=\dfrac{-18}{-6}
\\\\
x=3
.\end{array}
Upon checking, $
x=3
$ satisfies the original equation.
