jmc

We can factor the given equation as $$(2x+3)x^{2000}+(2x+3)x^{1998}+\cdots +(2x+3)=(2x+3)(x^{2000}+x^{1998}+\cdots +1)=0.$$Thus, $x=-\frac{3}{2}$ is a root. Note that $$x^{2000}+x^{1998}+\cdots +1\ge 1$$for all real $x,$ so the given polynomial has only $\boxed{1}$ real root.