## Precalculus (6th Edition) Blitzer

$\left\{4-\sqrt5, 4+\sqrt5\right\}$
Divide 3 on both sides of the equation to obtain: $(x-4)^2=5$ Take the square root of both sides to obtain: $x-4 = \pm \sqrt{5}$ Add 4 on both sides of the equation to obtain: $x=4 \pm \sqrt{5}$ Thus, the solution set of the given equation is $\left\{4-\sqrt5, 4+\sqrt5\right\}$.