$\begin{array}{l}
15 x^{2}-28=x \\
\Rightarrow 15 x^{2}-x-28=0 \\
\Rightarrow 15 x^{2}-(21 x-20 x)-28=0 \\
\Rightarrow 15 x^{2}-21 x+20 x-28=0
\end{array}$
$\begin{array}{l}
\Rightarrow 3 x(5 x-7)+4(5 x-7)=0 \\
\Rightarrow(3 x+4)(5 x-7)=0 \\
\Rightarrow 3 x+4=0 \text { or } 5 x-7=0 \\
\Rightarrow x=\frac{-4}{3} \text { or } x=\frac{7}{5}
\end{array}$

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