Biquadratic

We can rewrite the equation as $2(12x-1)4(6x-1)6(4x-1)8(3x-1) = 5\cdot2\cdot4\cdot6\cdot8$

$\Rightarrow (24x-2)(24x-4)(24x-6)(24x-8) = 1920$

Let $\quad$ $24x - 5 = a$

Then , our equation becomes $(a-3)(a-1)(a+1)(a+3) = 1920$

$\Rightarrow (a^2-1)(a^2-9) = 1920$

$\Rightarrow a^4-10a^2-1911=(a^2+39)(a^2-49)=0$

$\Rightarrow a^2=-39,49$

But since we want real solutions , we reject $a^2=-39$

$\Rightarrow a^2=49 \Rightarrow a = \pm 7 \Rightarrow 24x-5=\pm7$

$\Rightarrow \boxed{x=\dfrac12 , \dfrac{-1}{12}}$

---

Required answer is $\boxed{-24}$