#### Answer

$x_{1}=1 + \sqrt{6}$ and $x_{2} = 1 - \sqrt{6}$

#### Work Step by Step

Given $x(x-2)=5 \longrightarrow x^2-2x-5=0$
$a= 1, \ b=-2, \ c=-5$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-(-2) \pm \sqrt{(-2)^2-4 \times 1\times (-5)}}{2 \times 1} = \dfrac{2 \pm \sqrt{4+20}}{2} = \dfrac{2 \pm \sqrt{24}}{2} = \dfrac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6}$
Therefore the solutions are $x_{1}=1 + \sqrt{6}$ and $x_{2} = 1 - \sqrt{6}$