Looking at the root locus with the damping factor and natural frequency isolines, we can better see where the poles move.
For continuous-time systems, we have that the settling time is given by
$$ T_s = \frac{\ln(0.05\sqrt{1-\zeta^2})}{\zeta \omega_0},$$
As an approximation, to reach the smallest possible settling time you would want to maximize
$$ \max_{K,\alpha, \beta} \zeta \omega_0.$$
Or, as a proper minimization,
$$ \min_{K,\alpha, \beta} T_s = \min_{K,\alpha, \beta} \frac{\ln(0.05\sqrt{1-\zeta^2})}{\zeta \omega_0}.$$
I have marked a ballpark point where the settling time would be minimized, and by using a system where \$ C(z)=K \$ the best place is where they first break off of the real line.
By using a \$ C(z)=K\frac{z-\beta}{z-\alpha} \$, as you suggested, it is possible to (in theory) exactly cancel the pole \$ z_p = -0.2\$ and the zero \$ z_z = 0.5\$, then the system would become just
$$ \hat{G} = 2K\frac{z-2}{z+1.5},$$
And the feedback system would be
$$ H(z) = \frac{\hat{G}}{1+\hat{G}} = \frac{1}{\left(2K\frac{z-2}{z+1.5}\right)^{-1}+1} = \frac{2K(z-2)}{(z+1.5)+2K(z-2)},$$
$$ H(z) = \frac{2K(z-2)}{(2K+1)z+1.5-4K}.$$
We then find \$K\$ such that the only remaining pole is at zero, which is
$$(2K+1)z+1.5-4K = \alpha z + 0 \Leftrightarrow 1.5-4K = 0 \Leftrightarrow K = \frac{1.5}{4} = 0.375.$$
That way,
$$ H(z) = \frac{2 \cdot 0.375 (z-2)}{(2 \cdot 0.375 +1) z} = \frac{3}{7}\frac{(z-2)}{z} = \frac{3}{7}(1-2z^{-1}).$$
This way the settling time will be just \$T_s = 1\$, but do notice that the system is non-minimum phase ("goes in the wrong direction first" and has undershoot of 100%!)
