How do you solve #( x - 6)^2 = 19#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Meave60 Apr 14, 2016 #x=6+sqrt19, 6-sqrt 19# Explanation: #(x-6)^2=19# Take the square root of both sides. #x-6=+-sqrt 19# Add #6# to both sides of the equation. #x=6+-sqrt 19# Solve for #x#. #x=6+sqrt19, 6-sqrt 19# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 1800 views around the world You can reuse this answer Creative Commons License