We want to find the values of $b$ such that the quadratic $y=x^2+bx+12$ does not have $-4$ in its range.
To find the range of the quadratic, we can complete the square.
We have $y=(x+\frac{b}{2})^2+12-\frac{b^2}{4}$.
Since the square of a real number is always nonnegative, the minimum value of the quadratic occurs at $x=-\frac{b}{2}$,
and this minimum value is $12-\frac{b^2}{4}$.
So, for $-4$ to not be in the range of the quadratic, we need $12-\frac{b^2}{4}>-4$.
Simplifying, we get $b^2<64$, which means $-8<b<8$.
The greatest integer value of $b$ that satisfies this inequality is $\boxed{7}$.
The answer is: 7
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the greatest integer value of $b$ such that $-4$ is not in the range of $y=x^2+bx+12$?

### Answer:
