By comparing the given equation with standard equation $\mathrm{a}=10, \mathrm{b}=-27 \mathrm{and} \mathrm{c}=5.$

$\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$

$\Rightarrow \mathrm{x}=\frac{-(-27) \pm \sqrt{(-27{)}^2-4\times 10\times 5}}{2\times 10}=\frac{27\pm 23}{20}$

$\Rightarrow {\mathrm{x}}_{1 }=\frac{27+23}{20}=\frac{5}{2}\mathrm{and} x_2=\frac{27-23}{20}=\frac{1}{5}$

Roots of the equation are $\frac{5}{2}$ and $\frac{1}{5}$.