jmc

We complete the square.

Factoring $4$ out of the quadratic and linear terms gives $4x^2+2x=4\left(x^2+\frac{1}{2}x\right)$.

Since ${\left(x+\frac{1}{4}\right)}^2=x^2+\frac{1}{2}x+\frac{1}{16}$, we can write $$4{\left(x+\frac{1}{4}\right)}^2=4x^2+2x+\frac{1}{4}.$$This quadratic agrees with the given $4x^2+2x-1$ in all but the constant term. We can write

$$\begin{array}{rl}4x^2+2x-1& =\left(4x^2+2x+\frac{1}{4}\right)-\frac{5}{4}\\ & =4{\left(x+\frac{1}{4}\right)}^2-\frac{5}{4}.\end{array}$$Therefore, $a=4$, $b=\frac{1}{4}$, $c=-\frac{5}{4}$, and $a+b+c=4+\frac{1}{4}-\frac{5}{4}=\boxed{3}$.