The inequation is
#x^2+x-12<0#
Let's factorise the expression
#x^2+x-12=(x-3)(x+4)#
Let #f(x)=x^2+x-12#
Now, we can do the sign chart
#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##3##color(white)(aaaa)##-oo#
#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-3##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)<0#, when #x in ] -4,3 [#
or #-4 < x <3#