Solve $x+3<\sqrt{x+33}$ : Solution Verification First of all the part inside the square root should be greater than or equal to $0$. So, 
$x+33 \ge0$
$\Rightarrow x \ge -33$
Now I solved the inequality,
$x+3<\sqrt{x+33}$
$\Rightarrow(x+3)^2<{x+33}$
$\Rightarrow x^2+9+6x<{x+33}$
$\Rightarrow x^2+5x-24 < 0$
$\Rightarrow (x+8)(x-3) < 0$
$\Rightarrow -8<x<3$
Now combining the two results i.e. $ x \ge -33$ and $ -8<x<3$ , the answer should be $ -8<x<3$ as per my understanding but this is not the correct answer that has been provided. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!

HINT
You can only square both sides iff both sides and the argument of the square root are nonnegative.
Under such conditions, the proposed inequation is equivalent to
\begin{align*}
x + 3 < \sqrt{x + 33} & \Longleftrightarrow
\begin{cases}
(x+3)^{2} < x + 33\\\\
x+3 \geq 0\\\\
x + 33 \geq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x + 9 < x + 33\\\\
x\geq -3
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x - 34 < 0\\\\
x \geq -3
\end{cases}
\end{align*}
Moreover, notice whenever $-33 \leq x \leq -3$, the LHS is negative and the RHS is positive.
Hence the interval $[-33,-3]$ is also a solution.
Can you take it from here?
