Chapter 8 - Review - Page 531: 13

$x=8$

Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solution/s of the given quadratic equation, $ x^2-16x+64=0 ,$ is/are \begin{array}{l}\require{cancel} x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4(1)(64)}}{2(1)} \\\\ x=\dfrac{16\pm\sqrt{256-256}}{2} \\\\ x=\dfrac{16\pm\sqrt{0}}{2} \\\\ x=\dfrac{16\pm0}{2} \\\\ x=\dfrac{16}{2} \\\\ x=8 .\end{array}