What are the potential solutions to the equation below? 2in(x+3)=0

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!