First of all, remember that dividing by a fraction is the same as multiplying by the inverse of that fraction:
[tex]\dfrac{x^2-4}{x+3}\div\dfrac{x^2-4x+4}{4x+12}=\dfrac{x^2-4}{x+3}\cdot\dfrac{4x+12}{x^2-4x+4}[/tex]
Now, we can factor all numerators and denominators:
[tex]\dfrac{x^2-4}{x+3}\cdot\dfrac{4x+12}{x^2-4x+4}=\dfrac{(x+2)(x-2)}{x+3}\cdot\dfrac{4(x+3)}{(x-2)^2}[/tex]
and simplify whatever appears at both numerator and denominator:
[tex]\dfrac{(x+2)}{1}\cdot\dfrac{4}{(x-2)}=\dfrac{4(x+2)}{x-2}[/tex]
