How do you solve #(4x-1)^2= 16#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Babette Oct 29, 2015 #x=-3/4# and #x=5/4# Explanation: Take the square root of both sides #sqrt((4x-1)^2)=+-sqrt(16)# #4x-1=+-4# So we need to solve #4x-1=4# and #4x-1=-4# #4x=5# #x=5/4# #4x=-3# #x=-3/4# 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 2634 views around the world You can reuse this answer Creative Commons License