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.

As the logarithm term contains squares, we can allow $x-5,x+6$ to be negative
In fact, $\log b^2=2\log |b|$  for real $b$
As $\log A+\log B=\log (AB),$
$$\implies \log_{10}\{(x-5)(x+6)\}^2=2$$
$$\implies \{(x-5)(x+6)\}^2=10^2=100$$
$$\implies (x-5)(x+6)=\pm10$$
