(1) 2s2-6s-2=2(s2-3s-1)=2(s2-3s+9/4-9/4-1)=2((s-3/2)2-13/4).
This can be written 2(s-3/2-√13/2)(s-3/2+√13/2). 3s-s=2s.
s/((s-3/2-√13/2)(s-3/2+√13/2))=A/(s-3/2-√13/2)+B/(s-3/2+√13/2), where A and B are constants to be found.
A(s-3/2+√13/2)+B(s-3/2-√13/2)=s, so equating coefficients:
A+B=1, A(-3/2+√13/2)+B(-3/2-√13/2)=0=(A+B)(-3/2)+A√13/2-B√13/2,
(A-B)√13/2=3/2, A-B=3/√13. Therefore 2A=1+3/√13, A=(1+3/√13)/2=(13+3√13)/26 and B=(13-3√13)/26.
So (3s-s)/(2s2-6s-2)=(1/26)((13+3√13)/(s-3/2-√13/2)+(13-3√13)/(s-3/2+√13/2)).
ℒ-1{1/(s-a)}=eat, so if we let a=3/2+√13/2, b=3/2-√13/2, p=13+3√13 and q=13-3√13:
ℒ-1{(3s-s)/(2s2-6s-2)}=(peat+qebt)/26=
[(13+3√13)e(3/2+√13/2)t+(13-3√13)e(3/2-√13/2)t]/26=(e3/2/26)((13+3√13)et√13/2+(13-3√13)e-t√13/2).
(2) s2-3s-10=(s-5)(s+2),
(s+7)/(s2-3s-10)=A/(s-5)+B/(s+2),
A(s+2)+B(s-5)=s+7,
A+B=1, 2A-5B=7, 2A-5(1-A)=7, 2A-5+5A=7, 7A=12, A=12/7, B=-5/7.
(s+7)/(s2-3s-10)=(1/7)(12/(s-5)-5/(s+2)).
ℒ-1{(s+7)/(s2-3s-10)}=(1/7)(12e5t-5e-2t).
I suspect that the first given Laplace expression (1) has been wrongly presented because of the complexity that resulted in finding the inverse. The numerator 3s-s is too simplistic to be correct, and the quadratic denominator should have had rational factors. The Laplace expression in (2) was a lot simpler and uncomplicated.