#### Answer

$\dfrac {1\pm \sqrt {41}}{4}$

#### Work Step by Step

$\log _{2}\left( 2x-3\right) +\log _{2}\left( x+1\right) =1\Rightarrow \log _{2}\left( \left( 2x-3\right) \left( x+1\right) \right) =\log _{2}2$
$\Rightarrow \left( 2x-3\right) \left( x+1\right) =2\Rightarrow 2x^{2}-x-3=2\Rightarrow 2x^{2}-x-5=0$
$x_{1,2}=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\dfrac {1\pm \sqrt {1-4\times 2\times \left( -5\right) }}{2\times 2}=\dfrac {1\pm \sqrt {41}}{4}$