jmc

Let the third root be $s.$ Then $$8x^3-4x^2-42x+45=8(x-r{)}^2(x-s)=8x^3-8(2r+s)x^2+8(r^2+2rs)x-8r^{2}s.$$Matching coefficients, we get $$\begin{array}{rl}2r+s& =\frac{1}{2},\\ r^2+2rs& =-\frac{21}{4},\\ r^{2}s& =-\frac{45}{8}.\end{array}$$From the first equation, $s=\frac{1}{2}-2r.$ Substituting into the second equation, we get $$r^2+2r\left(\frac{1}{2}-2r\right)=-\frac{21}{4}.$$This simplifies to $12r^2-4r-21=0,$ which factors as $(2r-3)(6r+7)=0.$ Thus, $r=\frac{3}{2}$ or $r=-\frac{7}{6}.$

If $r=\frac{3}{2},$ then $s=-\frac{5}{2}.$ If $r=-\frac{7}{6},$ then $s=\frac{17}{6}.$ We can check that only $r=\boxed{\frac{3}{2}}$ and $s=-\frac{5}{2}$ satisfy $r^{2}s=-\frac{45}{8}.$