#### Answer

$x=0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{9-x}=x+3
,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation.
$\bf{\text{Solution Details:}}$
Squaring both sides of the equation results to
\begin{array}{l}\require{cancel}
\left( \sqrt{9-x} \right)^2=\left( x+3 \right)^2
\\\\
9-x=\left( x+3 \right)^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
9-x=(x)^2+2(x)(3)+(3)^2
\\\\
9-x=x^2+6x+9
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-x^2+(-x-6x)+(9-9)=0
\\\\
-x^2-7x=0
\\\\
-1(-x^2-7x)=-1(0)
\\\\
x^2+7x=0
.\end{array}
Using factoring by $GCF,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(x+7)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+7=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+7=0
\\\\
x=-7
.\end{array}
Upon checking, only $
x=0
$ satisfies the original equation.