Given: $\left|\begin{array}{ccc}2 \mathrm{x}+1 & 4 & 8 \\ 2 & 2 \mathrm{x} & 2 \\ 7 & 6 & 2 \mathrm{x}\end{array}\right|=0$
$\begin{aligned} \therefore(2 x+1)\left(4 x^{2}-12\right)-4 &(4 x-14) \\ &+8(12-14 x)=0 \end{aligned}$
$\begin{aligned} \Rightarrow 8 x^{3}-24 x+4 x^{2}-& 12-16 x \\ &+56+96-112 x=0 \end{aligned}$
$\Rightarrow 8 x^{3}+4 x^{2}-152 x+140=0$
$(x+5)$ is a factor of above equation
$\begin{aligned}
\therefore 8 x^{3}+40 x^{2}-36 x^{2}-180 x & \\
&+28 x+140=0 \\
\Rightarrow 8 x^{2}(x+5)-36 x(x+5) &+28(x+5)=0
\end{aligned}$
$\Rightarrow(x+5)\left(8 x^{2}-36 x+28\right)=0$
$\Rightarrow(x+5) 4\left(2 x^{2}-9 x+7\right)=0$
$\Rightarrow 4(x+5)\left(2 x^{2}-7 x-2 x+7\right)=0$
$\Rightarrow 4(x+5)[x(2 x-7)-1(2 x-7)]=0$
$\Rightarrow 4(x+5)(2 x-7)(x-1)=0$
$\Rightarrow x=-5,3.5,1$