As usual with absolute values, split into cases:
Case 1: #x^2 - 3 < 0#
If #x^2 - 3 < 0# then #abs(x^2-3) = -(x^2-3) = -x^2+3#
and our (corrected) inequality becomes:
#-x^2+3 < 2#
Add #x^2-2# to both sides to get #1 < x^2#
So #x in (-oo,-1) uu (1, oo)#
From the condition of the case we have
#x^2 < 3#, so #x in (-sqrt(3), sqrt(3))#
Hence:
#x in (-sqrt(3), sqrt(3)) nn ((-oo,-1) uu (1, oo))#
#= (-sqrt(3), -1) uu (1, sqrt(3))#
Case 2: #x^2 - 3 >= 0#
If #x^2 - 3 >= 0# then #abs(x^2-3) = x^2+3# and our (corrected) inequality becomes:
#x^2-3 < 2#
Add #3# to both sides to get:
#x^2 < 5#, so #x in (-sqrt(5), sqrt(5))#
From the condition of the case we have
#x^2 >= 3#, so #x in (-oo, -sqrt(3)] uu [sqrt(3), oo)#
Hence:
#x in ((-oo, -sqrt(3)] uu [sqrt(3), oo)) nn (-sqrt(5), sqrt(5))#
#= (-sqrt(5), -sqrt(3)] uu [sqrt(3), sqrt(5))#
Combined:
Putting case 1 and case 2 together we get:
#x in (-sqrt(5), -sqrt(3)] uu (-sqrt(3), -1) uu (1, sqrt(3)) uu [sqrt(3), sqrt(5))#
#=(-sqrt(5), -1) uu (1, sqrt(5))#