smc

Denote $s$ as the third solution. Then, by Vieta's, $2r+s=\frac{1}{2}$, $r^2+2rs=-\frac{21}{4}$, and $r^{2}s=-\frac{45}{8}$. Multiplying the top equation by $2r$ and eliminating, we have $3r^2=r+\frac{21}{4}$. Combined with the fact that $s=\frac{1}{2}-2r$, the third equation can be written as $(\frac{r+\frac{21}{4}}{3})(\frac{1}{2}-2r)=-\frac{45}{8}$, or $(4r+21)(4r-1)=135$. Solving, we get $r=\frac{3}{2},-\frac{13}{2}$. Plugging the solutions back in, we see that $-\frac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{1.52}$