#### Answer

$x_{1}= 2 + 3\sqrt{3}$ and $x_{2} = 2 - 3\sqrt{3}$

#### Work Step by Step

Given $(x-2)^2=27 \longrightarrow x^2-4x+4=27 \longrightarrow x^2-4x-23=0$
$a= 1, \ b=-4, \ c=-23$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-(-4) \pm \sqrt{(-4)^2-4 \times 1\times (-23)}}{2 \times 1} = \dfrac{4 \pm \sqrt{16+92}}{2} = \dfrac{4 \pm \sqrt{108}}{2} = \dfrac{4 \pm 6\sqrt{3}}{2} = \\ 2 \pm 3\sqrt{3}$
Therefore the solutions are $x_{1}= 2 + 3\sqrt{3}$ and $x_{2} = 2 - 3\sqrt{3}$