Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 22

$x=\{ -11,9 \} $

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ (x+1)^2=100 ,$ are \begin{array}{l}\require{cancel} x+1=\pm\sqrt{100} \\\\ x+1=\pm\sqrt{(10)^2} \\\\ x+1=\pm10 \\\\ x=-1\pm10 .\end{array} Hence, $ x=\{ -11,9 \} .$