How do you solve #(2x - 3)^2 = 1#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer ali ergin Apr 19, 2016 #x={1,2}# Explanation: #(2x-3)^2=1# #sqrt((2x-3)^2)=sqrt1# #2x-3=±1# #if +1# #2x=1+3# #2x=4# #x=4/2# #x=2# #if -1# #2x-3=-1# #2x=2# #x=1# 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 1119 views around the world You can reuse this answer Creative Commons License