Find the number of real solutions of the equation
\[\frac{4x}{x^2 + x + 3} + \frac{5x}{x^2 - 5x + 3} = -\frac{3}{2}.\]
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.