#### Answer

$x_{1}=5;x_{2}=-20$

#### Work Step by Step

$\log x+\log \left( x+15\right) =2\Rightarrow \log _{10}\left( x\left( x+15\right) \right) =2$
$x\left( x+15\right) =10^{2}\Rightarrow x^{2}+15x-100=0$
$x_{1,2}=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\dfrac {-15\pm \sqrt {15^{2}-4\times 1\times \left( -100\right) }}{2}=\dfrac {-15\pm 25}{2}=-20; 5$