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?