This might be a silly question to ask but I encountered a problem during solving this question. The question says to find the domain of $f(x) = \sqrt{-3x^2-5x+8}$. For the function $f(x)$ to have real values, the quantity $-3x^2-5x+8 \ge 0$. $$ -3x^2+3x-8x+8 \ge 0 \\ 3x(-x+1)+8(-x+1) \ge 0 \\ (3x+8)(-x+1) \ge 0 \\ x \ge \frac{-8}{3} \quad \text{or}\quad x\le 1 $$ Well, the domain of the is calculated to be $[\frac{-8}{3},1]$. Plotting the function on graph, we can be further be assured that the domain I have calculated is correct. But, when I tried solving this again and encountered exactly opposite domain.
$$ -3x^2+3x-8x+8 \ge 0 \\ -3x(x-1)-8(x-1) \ge 0 \\ (-3x-8)(x-1) \ge 0 \\ x \le \frac{-8}{3} \quad \text{or}\quad x\ge 1 $$
During the second process, I took $-3x$ and $-8$ as common factor. After that, I simplified and got the domain of the function to be $(-\infty, \frac{-8}{3}] \cup [1,\infty)$. I got two different answers just by changing the factor to negative. I think I am missing something crucial here and making a mistake. Please help me find the reason what and why am I doing wrong.
Note that Dom($f)=\{x\in\mathbb{R}~|~-3x^2-5x+8\ge 0\}$, where $f(x)=-3x^2-5x+8$.
Now, $-3x^2-5x+8\ge 0\iff (3x+8)(x-1)\le 0\iff 3x+8 \text{ and }x-1 \text{ have different signs}$. So it is enough to investigate the following two possibilities:
(i) $3x+8\ge 0$ and $x-1\le 0$, i.e., $x\in\left[-\frac{8}{3},1\right]$, or
(ii) $3x+8\le 0$ and $x-1\ge 0$, which is of course absurd.
Thus Dom($f$)=$\left[-\frac{8}{3},1\right]$.