jmc

Since the parabola is completely below the $x$-axis, it must open downwards (otherwise, it would have to cross the $x$ axis on its way up). This means $a<0$. We have $a^2=49$ so $a=\pm 7$, but since $a$ is negative $a=-7$.

Since our graph does not touch the $x$-axis, we must not have any real solutions. Since all solutions must be imaginary, the discriminant must be negative, or $$\begin{array}{rl}{b}^2-4ac& <0\Rightarrow \\ b^2-4(-7)(-6)& <0\Rightarrow \\ b^2-168& <0\Rightarrow \\ b^2& <168.\end{array}$$ This means that $-\sqrt{168}<b<\sqrt{168}$. The largest integral value of $b$ is the greatest integer less than $\sqrt{168}$. Since $13^2=169$, we know $\sqrt{168}$ is a little less than $13$ but more than $12$. So the greatest integral value of $b$ is $\boxed{12}$.