Step 1: Find the roots of the given quadratic.
Solve $2x^2-x-6=0$. Factor as $(x-2)(2x+3)=0$, so the roots are $2$ and $-\dfrac32$.
Step 2: Name them in order.
Since $\alpha<\beta$, we take $\alpha=-\dfrac32$ and $\beta=2$.
Step 3: Write the inequality clearly.
We need $\alpha x^2+kx-\beta\le0$ for all real $x$, that is $-\dfrac32 x^2+kx-2\le0$ always.
Step 4: Check the shape.
The leading coefficient $-\dfrac32$ is negative, so the parabola opens downward. A downward parabola stays at or below zero everywhere only when it never crosses the axis.
Step 5: Use the discriminant condition.
We need discriminant $\le0$: $k^2-4\left(-\dfrac32\right)(-2)=k^2-12\le0$, so $k^2\le12$, giving $-2\sqrt3\le k\le2\sqrt3$.
Step 6: Count whole numbers.
Since $2\sqrt3\approx3.46$, the integers are $-3,-2,-1,0,1,2,3$, which is $7$ values. \[ \boxed{7} \]