Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=\pm10$
Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), then the solutions to the given equation, $x^2=100 ,$ are \begin{array}{l}\require{cancel} x=\pm\sqrt{100} \\\\ x=\pm\sqrt{(10)^2} \\\\ x=\pm10 .\end{array}