We write, $6 x=x+5 x$ as $x^{2} \times 5=5 x^{2}=x \times 5 x$

$\begin{array}{l}
\therefore x^{2}+6 x+5=0 \\
\Rightarrow x^{2}+x-5 x+5=0
\end{array}$
$\begin{array}{l}
\Rightarrow x(x+1)+5(x+1)=0 \\
\Rightarrow(x+1)(x+5)=0 \\
\Rightarrow x+1=0 \text { or } x+5=0 \\
\Rightarrow x=-1 \text { or } x=-5
\end{array}$

Hence, the roots of the given equation are $-1$ and $-5$.