$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:

$$\sqrt{9x^2-16}>3x+1$$


Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.

Where, what, and why is wrong with my solution?

at first it must be $|x|\geq \frac{4}{3}$ if $$x\le -\frac{1}{3}$$ our inequality is true. let $$x>-\frac{1}{3}$$ we can square it:
$$9x^2-16>9x^2+6x+1$$ otr $$x<-\frac{17}{6}$$ together with the conditions $$-\infty<x<-\frac{17}{6}$$ and $$-\frac{1}{3}<x$$ or $$-\frac{1}{3}\geq x$$ and $$x\geq \frac{4}{3}$$ or $$x\le -\frac{4}{3}$$ we get $$-\infty<x\le- \frac{4}{3}$$ as the searched solution set
