logarithms equation Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$ 
and I rewrited this equation as 
$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $
and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\log{(x-5)^2(x+6)^2}=\log{100}$ or is a simplier way.

$$\log{(x-5)^2}+\log{(x+6)^2}=2$$
$$\log{((x-5)(x+6))^2}=2$$
$$((x-5)(x+6))^2=10^2=100$$
$$(x-5)(x+6)=\pm10$$Solve from here
