If this is the full question; What are the potential solutions to the equation below? 2In(x+3)=0
then the answer is; 2ln(x + 3) = 0
ln[(x + 3)²] = 0
(x + 3)² = 1
x + 3 = ±√1
x + 3 = ±1
x = 1 - 3, -1 - 3
x = -2, -1 - 3
x = -2, -4
when checking solution; x = -4 in the original equation does not hold true. so you drop x = -4 from the solution set.
therefore;
x = -2
hope this helps, God bless!