Variable $x$ cannot be equal to $-3$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x+3\right)$, the least common multiple of $2x+6,2$.
$$x^{2}-9=\left(x+3\right)\left(x-3\right)$$
Consider $\left(x+3\right)\left(x-3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.
