Bi-quadratics aren't easy

We can multiply the first term with the fourth term, and the second one with the third one:

$(36x^2-15x+1)(24x^2-10x+1)=5$

$(3(12x^2-5x)+1)(2(12x^2-5x)+1)=5$

Now, let $y=12x^2-5x$ , so:

$(3y+1)(2y+1)=5 \\ 6y^2+5y-4=0 \\ (3y+4)(2y-1)=0 \\ y_1=-\dfrac{4}{3} \\ y_2=\dfrac{1}{2}$

From here we have two solutions for $y$ , so there are two cases:

$12x^2-5x=-\dfrac{4}{3} \\ 36x^2-15x+4=0$

The discriminant is $\Delta=(-15)^2-4(36)(4)=-351<0$ , hence we don't have real roots.

The second case is:

$12x^2-5x=\dfrac{1}{2} \\ 24x^2-10x-1=0$

The discriminant is $\Delta=(-10)^2-4(24)(-1)=196>0$ , so there are two distinct real solutions, and their product, by Vieta's formulas is $-\dfrac{1}{24}$ .

Comparing we get $a=-1$ and $b=24$ , thus the answer is $|a|+b=\boxed{25}$ .