jmc

First, we rearrange the equation so that one side is a quadratic written in the usual way and the other side is $0$. We can do this by subtracting $20x$ from each side (and reordering the terms): $$5x^2-20x+18=0$$This doesn't factor in any obvious way, so we apply the quadratic formula, which gives $$\begin{array}{rl}x=\frac{-(-20)\pm \sqrt{(-20{)}^2-4(5)(18)}}{2(5)}& =\frac{20\pm \sqrt{400-360}}{10}\\ & =\frac{20\pm \sqrt{40}}{10}\\ & =2\pm \frac{\sqrt{40}}{10}.\end{array}$$We observe that $\sqrt{40}$ is between $6$ and $7$, so $\frac{\sqrt{40}}{10}$ is between $0.6$ and $0.7$. Thus one of our solutions is between $1.3$ and $1.4$, while the other is between $2.6$ and $2.7$. Rounding each solution to the nearest integer, we get $1$ and $3$, the product of which is $\boxed{3}$.