jmc

Let $y=x^2+x+3.$ Then we can write the given equation as $$\frac{4x}{y}+\frac{5x}{y-6x}+\frac{3}{2}=0.$$Multiplying everything by $2y(y-6x),$ we get $$8x(y-6x)+10xy+3y(y-6x)=0.$$Expanding, we get $3y^2-48x^2=0,$ so $y^2-16x^2=(y-4x)(y+4x)=0.$ Thus, $y=4x$ or $y=-4x.$

If $y=4x,$ then $x^2+x+3=4x,$ so $x^2-3x+3=0.$ This quadratic has no real solutions.

If $y=-4x,$ then $x^2+x+3=-4x,$ so $x^2+5x+3=0.$ This quadratic has two real solutions, giving us a total of $\boxed{2}$ real solutions.