jmc

We proceed as follows: $$\begin{array}{rl}6a^2+5a+4& =3\\ 6a^2+5a+1& =0\\ (2a+1)(3a+1)& =0.\end{array}$$This gives us $a=-\frac{1}{2}$ or $a=-\frac{1}{3}.$ Of these, $a=-\frac{1}{2}$ gives the smaller value of $2a+1=\boxed{0}.$