It's neither an arithmetic sequence (no common difference) nor a geometric sequence (no common ratio). It is nevertheless possible to find a function that will enable the next two terms to be calculated. Since 4 terms are given there should be a cubic function f(n)=an3+bn2+cn+d where coefficients a, b, c, d are numbers to be found and 0≤n≤3, so f(0)=-3.6, f(1)=-5.6, f(2)=-8.1, f(3)=-12.15.
f(0)=d so d=-3.6.
f(1)+3.6=a+b+c, a+b+c=-5.6+3.6=-2 (1)
f(2)+3.6=8a+4b+2c, 8a+4b+2c=-8.1+3.6=-4.5 (2)
f(3)+3.6=27a+9b+3c, 27a+9b+3c=-12.15+3.6=-8.55 (3)
The parenthesised numbers are labels for the equations.
We have three equations and three unknowns. Now we can manipulate the equations to eliminate one coefficient at a time.
(4)=(2)-2(1): 6a+2b=-4.5+4=-0.5
(5)=(3)-3(1): 24a+6b=-8.55+6=-2.55
(6)=(5)-3(4): 6a=-2.55+1.5=-1.05, a=-1.05/6=-0.175.
Substituting for a in (4): -1.05+2b=-0.5, 2b=1.05-0.5=0.55, b=0.275.
Substituting for a and b in (1): -0.175+0.275+c=-2, c=-2+0.175-0.275=-2.1
f(n)=-0.175n3+0.275n2-2.1n-3.6.
The series becomes -3.6, -5.6, -8.1, -12.15, -18.8, -29.1, ...
Note that this is one solution if the series is a simple polynomial. There will be other solutions which are not simple polynomials.