Solving $\sqrt{8+2x-x^2} > 6-3x$. 
So I was solving this inequality.
  $$\sqrt{8+2x-x^2} > 6-3x$$



*

*First things first, I obtained that the common domain of definition is actually $[-2,4]$.

*Next we would square and solve the quadratic that follows.
But the "solution" seems to have a part, where they took $6-3x \geq 0$,
which gave another restriction for $x$ as $(-\infty,2]$. 
I did not understand this. Why was this necessary?

Just use these fundamental rules for irrational (in)equalities, valid on the domain $A\ge 0$:
\begin{alignat}{2}
&\bullet\quad \sqrt A> B\iff A>B^2\quad\text{OR}&&\quad B<0 \\
&\bullet\quad \sqrt A < B\iff A < B^2\quad\text{AND}&&\quad B\ge 0 \\
&\bullet\quad \sqrt A = B\iff A = B^2\quad\text{AND}&&\quad B\ge 0
\end{alignat}
