How do you solve #(x – 2) ² = 4(4 – x)#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Mar 27, 2016 #x=-2sqrt3# or #x=2sqrt3# Explanation: Simplifying #(x-2)^2=4(4-x)#, we get #x^2-4x+4=16-4x# or #x^2-4x+4x+4-16=0# or #x^2-12=0# or #(x+sqrt12)(x-sqrt12)=0# or #x=-sqrt12=-2sqrt3# or #x=2sqrt3# 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 6622 views around the world You can reuse this answer Creative Commons License