Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 24

$y=4\pm3\sqrt{2}$

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ (y-4)^2=18 ,$ are \begin{array}{l}\require{cancel} y-4=\pm\sqrt{18} \\\\ y-4=\pm\sqrt{9\cdot2} \\\\ y-4=\pm\sqrt{(3)^2\cdot2} \\\\ y-4=\pm3\sqrt{2} \\\\ y=4\pm3\sqrt{2} .\end{array}