jmc

The first quadratic factors as $$(2x-5)(x+11)=0,$$so its roots are $\frac{5}{2}$ and $-11.$ Since $\lfloor c\rfloor$ must be an integer, it must be the case that $\lfloor c\rfloor =-11.$

The second quadratic factors as $$(3x-1)(2x-7)=0,$$so its roots are $\frac{1}{3}$ and $\frac{7}{2}.$ Since $0\le \{c\}<1,$ it must be the case that $\{c\}=\frac{1}{3}.$

Then $c=\lfloor c\rfloor +\{c\}=-11+\frac{1}{3}=\boxed{-\frac{32}{3}}.$