Note 6x2+28x+30=2(3x+5)(x+3),
x2+6x+9=(x+3)2,
6x+10=2(3x+5).
Let a=log(x+3)[2(3x+5)]. Then LHS =a+1 and log2(3x+5)(x+3)2=a2.
Equation becomes a+1=5−a4, i.e., a2−4a+4=0⇒a=2.
So log(x+3)[2(3x+5)]=2⇒(x+3)2=6x+10⇒x2=1⇒x=1 or x=−1.
Both satisfy domain conditions (bases >0, =1; arguments >0).
Sum =1+(−1)=0.