Prove that

Given:

$15 x^{2}-28=x$

$\Rightarrow 15 x^{2}-x-28=0$

$\Rightarrow 15 x^{2}-(21 x-20 x)-28=0$

$\Rightarrow 15 x^{2}-21 x+20 x-28=0$

$\Rightarrow 3 x(5 x-7)+4(5 x-7)=0$

$\Rightarrow(3 x+4)(5 x-7)=0$

$\Rightarrow 3 x+4=0$ or $5 x-7=0$

$\Rightarrow x=\frac{-4}{3}$ or $x=\frac{7}{5}$

Hence, the roots of the equation are $\frac{-4}{3}$ and $\frac{7}{5}$.