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

hint: Use $|A| > |B| \iff A^2 > B^2 \iff (A-B)(A+B) > 0$. Apply this property to $A = x^2-2x-3, B = x^2+7x-13$
