$4 \sqrt{6} x^{2}-13 x-2 \sqrt{6}=0$
Given:
$4 \sqrt{6} \mathrm{x}^{2}-13 \mathrm{x}-2 \sqrt{6}=0$
$\Rightarrow 4 \sqrt{6} \mathrm{x}^{2}-16 \mathrm{x}+3 \mathrm{x}-2 \sqrt{6}=0$
$\Rightarrow 4 \sqrt{2} \mathrm{x}(\sqrt{3} \mathrm{x}-2 \sqrt{2})+\sqrt{3}(\sqrt{3} \mathrm{x}-2 \sqrt{2})=0$
$\Rightarrow(4 \sqrt{2} \mathrm{x}+\sqrt{3})(\sqrt{3} \mathrm{x}-2 \sqrt{2})=0$
$\Rightarrow 4 \sqrt{2} \mathrm{x}+\sqrt{3}=0$ or $\sqrt{3} \mathrm{x}-2 \sqrt{2}=0$
$\Rightarrow \mathrm{x}=\frac{-\sqrt{3}}{4 \sqrt{2}}=\frac{-\sqrt{3} \times \sqrt{2}}{4 \sqrt{2} \times \sqrt{2}}=\frac{-\sqrt{6}}{8}$ or $\mathrm{x}=\frac{2 \sqrt{2}}{\sqrt{3}}=\frac{2 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{2 \sqrt{6}}{3}$
Hence, the roots of the equation are $\frac{-\sqrt{6}}{8}$ and $\frac{2 \sqrt{6}}{3}$.