$|28 - x^2| < |3x|$
$\Rightarrow |x^2-28| < |3x| $
Case $1$ when $3x \geq 0 \Rightarrow x \geq0$ :-
$|x^2-28| < 3x$
$\Rightarrow -3x < x^2-28 < 3x $
$\Rightarrow x \in (4,7) \Rightarrow x=5,6$
Case $1$ when $3x \lt 0 \Rightarrow x \lt 0$ :-
$|x^2-28| < -3x$
$\Rightarrow -3x < x^2-28 < 3x $
$\Rightarrow x \in ( -\infty ,-7) \bigcup ( -4 ,0) $
How do we find the integers in the second case? Please help !!!
Thanks in advance !!!
There is an error when you wrote $$-3x<x^2-28<3x$$ Since you took $x<0$, $3x<-3x$; the correct inequality will be, $$3x<x^2-28<-3x$$ which gives $-7<x<-4$, giving the integers $x=-5,-6$.
Nonetheless, as mentioned in the comments by @RyszardSzwarc, you do not need to consider separate cases because the inequality remains the same if you replace $x$ by $-x$, so you can just fix one sign for $x$, and the negatives of those will also be a solution. (In this case, $5$ and $6$ being solutions implies $-5$ and $-6$ also being solutions.)
Hope this helps. :)