jmc

Multiplying the whole equation by $6x$ will get rid of the fractions: $$\begin{array}{rl}5x^2+6x& =18⟹\\ 5x^2+6x-18& =0.\end{array}$$Since the expression on the left-hand side does not factor easily, we use the quadratic formula to get $$\begin{array}{rl}x& =\frac{-6\pm \sqrt{36+360}}{10}\\ & =\frac{-6\pm \sqrt{396}}{10}\\ & =\frac{-6\pm 6\sqrt{11}}{10}.\end{array}$$Therefore, the largest possible value for $x$ is $\frac{-6+6\sqrt{11}}{10}$, or $\frac{-3+3\sqrt{11}}{5}$. Applying this to $\frac{a+b\sqrt{c}}{d}$, $a=-3$, $b=3$, $c=11$, and $d=5$. $$\frac{acd}{b}=\frac{-3\cdot 11\cdot 5}{3}=\boxed{-55}.$$