Let's rewrite the inequality as
#2x^3+5x^2-6x-9>0#
We must factorise the LHS
Let #f(x)=2x^3+5x^2-6x-9#
We calculate
#f(-1)=-2+5+6-9=0#
Therefore, #(x+1)# is a factor
To find the other factors, we do a long division
#color(white)(aaaa)##2x^3+5x^2-6x-9##color(white)(aaaa)##∣##x+1#
#color(white)(aaaa)##2x^3+2x^2##color(white)(aaaaaaaaaaaa)##∣##2x^2+3x-9#
#color(white)(aaaaaa)##0+3x^2-6x#
#color(white)(aaaaaaaa)##+3x^2+3x#
#color(white)(aaaaaaaaaa)##+0-9x-9#
#color(white)(aaaaaaaaaaaaaa)##-9x-9#
#color(white)(aaaaaaaaaaaaaaa)##-0-0#
Therefore,
#(2x^3+5x^2-6x-9)/(x+1)=2x^2+3x-9=(2x-3)(x+3)#
So,
#2x^3+5x^2-6x-9=(x+1)(2x-3)(x+3)>0#
Now we can do the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaa)##-1##color(white)(aaaa)##3/2##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##2x-3##color(white)(aaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)>0# when #x in ] -3, -1 [ uu ]3/2, +oo[#