The numerator of the left hand side can be rewritten as:
[tex]x^2+6x+9=(x+3)^2[/tex]Then, the given equation can be written as:
[tex]\frac{(x+3)^2}{x+3}=0[/tex]Since
[tex](x+3)^2=(x+3)(x+3)[/tex]we have
[tex]\frac{(x+3)(x+3)}{x+3}=0[/tex]We can to can cancel out one term x+3 and get
[tex]x+3=0[/tex]which gives
[tex]x=-3[/tex]Finally, in order to check that this value corresponds to a real answer, we need to subsitute this value into the equation and compute the limit when x approaches to -3, that is,
[tex]lim_{x\rightarrow-3}\frac{x^2+6x+9}{x+3}[/tex]which gives
[tex]l\imaginaryI m_{x\operatorname{\rightarrow}-3}\frac{x^{2}+6x+9}{x+3}=\frac{0}{0}[/tex]Since the limit has the form 0/0 we can to apply L'Hopital rule, that is,
[tex]l\imaginaryI m_{x\operatorname{\rightarrow}-3}\frac{\frac{d}{dx}(x^2+6x+9)}{\frac{d}{dx}(x+3)}[/tex]which gives
[tex]l\imaginaryI m_{x\operatorname{\rightarrow}-3}\frac{\frac{d}{dx}(x^{2}+6x+9)}{\frac{d}{dx}(x+3)}=l\imaginaryI m_{x\operatorname{\rightarrow}-3}\frac{2x+6}{1}=\frac{0}{1}=0[/tex]Since the limit exists and is equal to zero then the solution of the equation is: x= -3