Determine the values of $\text{'}m\text{'}$ for which the equation $5x^2-4x+2+m(4x^2-2x-1)=0$  when roots are equal 

$\begin{array}{l}5x^2-4x+2+4m{x}^2-2mx-m=0\\ (5+4m)x^2-(4+2m)x+2-m=0\\ a=5+4m,b=-(4+2m),c=2-m\end{array}$

When roots are equal

$\begin{array}{l}{b}^2-4ac=0\\ {(-(4+2m))}^2-4(5+4m)(2-m)=0\\ 16+4m^2+16m-4[10-5m+8m-4m^2]=0\end{array}$

$\begin{array}{l}16+4m^2+16m-40+20m-32m+16m^2=0\\ 20m^2+36m-32m-24=0\\ 20m^2+4m-24=0\end{array}$

$\begin{array}{l}5m^2+m-6=0\\ 5m^2+6m-5m-6=0\\ m[5m+6]-1[5m+6]=0\\ (m-1)(5m+6)=0\end{array}$

$m=1$ or $m=\frac{-6}{5}$