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))#