Solving $\sqrt{8+2x-x^2} > 6-3x$.

Let's say we have an inequality $\sqrt a>b$. We're often interested in getting rid of the square root, so we want to do something along the lines of 'squaring both sides'. But squaring both sides doesn't necessarily preserve the inequality.

Example:

$$\sqrt5>-3$$

But

$$\implies\sqrt5^2>(-3)^2 \implies 5\gt9$$

is clearly false.

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If you want a general rule, then you should use

$$|a|\gt|b|\iff a^2>b^2$$

This explains why you need to consider separate cases here

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}

The domain gives $-2\leq x\leq4$.

For $x>2$ with the domain our inequality is true.

But for $x\leq2$ we can use squaring,

which gives $5x^2-19x+14<0$ or $1<x<2.8$ and since $(1,2]\cup(2,4]=(1,4]$, we get the answer: $$(1,4]$$