How do you solve the equation #x^2 - 4x = 4#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Tom May 10, 2015 #x^2-4x=4# #=>x^2-4x+4=8# #a = x# #b = 2# #a^2-2ab+b^2 = (a-b)^2# #(x-2)^2=8# #x-2 = sqrt(8)# or #-sqrt(8)# #x_1 = sqrt(8)+2# #x_2= -sqrt(8)+2# 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 2079 views around the world You can reuse this answer Creative Commons License