How do you solve #(x - 4)^2 = 16#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nam D. Mar 29, 2018 #x=0,8# Explanation: Given: #(x-4)^2=16# Square root both sides: #x-4=sqrt(16)# #x-4=+-4# So we got: #x_1-4=4,x_2-4=-4#. #x_1=4+4# #=8# #x_2=-4+4# #=0# #:.x\in{0,8}# 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 10361 views around the world You can reuse this answer Creative Commons License