So I can rewrite the equation as $5x(x + 2y) = x^2(x + 2y).$ Now, if $x + 2y$ is not zero, I can divide both sides by it and get $5x = x^2.$ This is a quadratic equation that I can solve by moving everything to one side and factoring: $x^2 - 5x = 0 \implies x(x - 5) = 0.$ So $x = 0$ or $x = 5.$ But I remember that the problem said $x$ is a positive real number, so I can eliminate $x = 0$ and conclude that $x = 5.$ 
If $x$ and $y$ are positive real numbers such that $5x^2 + 10xy = x^3 + 2x^2 y,$ what is the value of $x$?