jmc

First, we move all terms to one side to get $4x^2-3x+12=0.$ Seeing that factoring will not work, we apply the Quadratic Formula: $$\begin{array}{rl}x& =\frac{-(-3)\pm \sqrt{(-3{)}^2-4(4)(12)}}{2(4)}\\ & =\frac{3\pm \sqrt{9-192}}{8}=\frac{3\pm \sqrt{-183}}{8}=\frac{3}{8}\pm \frac{\sqrt{183}}{8}i.\end{array}$$Now we see that $a=\frac{3}{8}$ and $b=\pm \frac{\sqrt{183}}{8},$ so $a+b^2=\frac{3}{8}+\frac{183}{64}=\boxed{\frac{207}{64}}.$