At first sight of the denominator on the right, g^2-16, we know that g=4 and g=-4 are not admissible.
To confirm, cross-multiply,
3(g^2-16)=(g+2)(2g+8)
Expand and simplify,
3g^2-48=2g^2+12g+16
g^2-12g-64=0
Solve by factoring,
(g-16)(g+4)=0
so g=16 or g=-4.
Since g=-4 will give a zero denominator on the right-hand side, it is inadmissible, and hence extraneous.