Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$ Here I have a question:
Solve for real value of $x$:
$$|x^2 -2x -3| > |x^2 +7x -13|$$
I got the answer as $x = (-\infty, \frac{1}{4}(-5-3\sqrt{17}))$ and $x=(\frac{10}{9},\frac{1}{4}(3\sqrt{17}-5)$
Please verify it if it is correct or not. Thanks

Brute force (as I am wont to do):
Look at where $(x^2-2x-3)^2 - (x^2+7x-13)^2 = -(9x-10)(2x^2+5x-16)$ is strictly positive.
The zeroes are ${10 \over 9}$ and ${1 \over 4} (-5 \pm 3 \sqrt{17})$.
Since the leading coefficient is $-1$, we see that the answer is
$(-\infty,{1 \over 4} (-5 - 3 \sqrt{17})) \cup ({10 \over 9}, {1 \over 4} (-5 + 3 \sqrt{17}))$.
