If [y] denote the greatest integer $< y,$and $2{\left[\frac{x}{8}\right]}^2+3\left[\frac{x}{8}\right]=20$ then $x$ lies in the smallest interval $[a, b)$ where$b– a$ is equal to 

$2{\left[\frac{x}{8}\right]}^2+3\left[\frac{x}{8}\right]-20=0$

$\Rightarrow \left[\frac{x}{8}\right]=\frac{5}{2}\text{ or }-4$

As $\left[\frac{x}{8}\right]$ is an integer, we take

$\left[\frac{x}{8}\right]=-4\Rightarrow -4\le x/8<-3$

$\Rightarrow -32\le x<-24$ Thus, $a = – 32,$ and $b = – 24$. Therefore,  $b – a = 8$