How do you solve #x^2-5x+4=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Azhar A. Apr 9, 2016 #x=4# or #x=1# Explanation: #x^2-5x+4=0# We need to factorize it: #x^2-x-4x+4=0# #x(x+1)-4(x-1)=0# #(x-4)(x-1)=0# #x-4=0# or #x-1=0# #x=4# or #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 4948 views around the world You can reuse this answer Creative Commons License