jmc

We have $p(q(x))=p(8x^2+10x-3)=\sqrt{-(8x^2+10x-3)}=\sqrt{-8x^2-10x+3}$. The input of this function is restricted since the quantity inside the square root cannot be negative. So we have $$\begin{array}{rl}-8x^2-10x+3& \ge 0\\ 8x^2+10x-3& \le 0\end{array}$$Factoring by trial and error gives $$(4x-1)(2x+3)\le 0$$Thus the roots of $8x^2+10x-3$ are $\frac{1}{4}$ and $-\frac{3}{2}$. Since we know the function $8x^2+10x-3$ is a parabola that opens up, its value is negative between the roots. Thus, our inequality is satisfied when $-\frac{3}{2}\le x\le \frac{1}{4}$. Thus $a=-\frac{3}{2}$, $b=\frac{1}{4}$, and $b-a=\frac{1}{4}-\left(-\frac{3}{2}\right)=\frac{1}{4}+\frac{6}{4}=\boxed{\frac{7}{4}}$.