Solving inequalities involving square roots What is the correct way to solve inequalities like $\sqrt {x+2}&gt;x $   ?

I deduced that $x$ must be greater than$-2$ and also on squaring I get $-1 \leq x \leq 2$ but even on combining these two I'm not getting the correct answer.What am I missing?
First note that $x$ must be greater than $-2$, because of the square root. Then you have two cases :

If $x \in [-2,0[$ the inequality is true since a square root is positive.

If $x \in [0,+\infty)$, then both sides are positive and you can square it to obtain the equivalent inequation $$x+2 &gt; x^2.$$ That is $$x^2-x-2 &lt; 0.$$ The discriminant here is $9$ and you can hence deduce the factorisation $$(x-2)(x+1) &lt;0.$$ Then do a sign table with the two factors to conclude.
Use: $$\sqrt{f(x)}&gt;g(x) \Leftrightarrow$$

$\Leftrightarrow \begin{cases}
g(x)&lt;0\\
f(x)\ge0\end{cases}$
or
$\begin{cases}
g(x)\ge0\\
f(x)&gt;\left(g(x)\right)^2
\end{cases}$


  !$$\sqrt{x+2}&gt;x$$
  
  !Case 1) $$\begin{cases}
x&lt;0\\
x+2\ge0\end{cases}$$
  
  !Case 2) $$\begin{cases}
x\ge0\\
x+2&gt;\left(x \right)^2
\end{cases}$$