Answer
$\frac{1\pm\sqrt {13}}{2}\approx- 1.303, 2.303$.
Work Step by Step
1. $log(x^2+3)=log(x+6) \Longrightarrow x^2+3=x+6 \Longrightarrow x^2-x-3=0 \Longrightarrow x=\frac{1\pm\sqrt {1-4(-3)}}{2}=\frac{1\pm\sqrt {13}}{2}\approx- 1.303, 2.303$.
2. Check, $x=\frac{1\pm\sqrt {13}}{2}$ fit(s).
