#### Answer

$x=\left\{ -2,-1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\sqrt{x+2}-x=2
,$ isolate the radical expression. Then square both sides and express in the form $ax^2+bx+c=0.$ Use concepts of solving quadratic equations to find the values of $x$. Finally, do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{x+2}=x+2
.\end{array}
Squaring both sides of the given equation results to
\begin{array}{l}\require{cancel}
x+2=(x+2)^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 equation above is equivalent to
\begin{array}{l}\require{cancel}
x+2=(x)^2+2(x)(2)+(2)^2
\\\\
x+2=x^2+4x+4
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-x^2+(x-4x)+(2-4)=0
\\\\
-x^2-3x-2=0
\\\\
-1(-x^2-3x-2)=-1(0)
\\\\
x^2+3x+2=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(x+2)(x+1)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
x+2=0
\\\\\text{OR}\\\\
x+1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+2=0
\\\\
x=-2
\\\\\text{OR}\\\\
x+1=0
\\\\
x=-1
.\end{array}
Upon checking, $
x=\left\{ -2,-1 \right\}
$ satisfy the original equation.