Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 592: 14

{$4+\sqrt{3},4-\sqrt{3}$}

Given: $(x-4)^2=3$ Apply The Square Root Property. This can be written as: $(x-4)=\sqrt{3}$ or, $(x-4)=\pm \sqrt{3}$ or, $x=4+\sqrt{3},4-\sqrt{3}$ Hence, solution set is $x=${$4+\sqrt{3},4-\sqrt{3}$}