$$d^{p-q}=\left (\frac32\right)^3 \tag{1}$$ $$ d^{r-q}=\left (\frac32\right) \tag{2}$$ $$ d^{p-r}=\left (\frac32\right)^2 \tag{3}$$ Clearly, $d=\frac32$ and $p=4,q=1,r=2$ so now our equation becomes $$ 4x^2+2x-4=0$$ We can use quadratic formula for the rest, if my algebra is correct the roots are $\frac{1}{4} \left(-1-\sqrt{17}\right)$ and $ \frac{1}{4} \left(-1+\sqrt{17}\right)$ of which only the later lies in $(0,1)$.