How do you solve #x^2 - 5x =0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 2 Answers Kushagra May 14, 2018 Factorize #x# as common #x(x-5)=0# Divide both sides by #x# #x-5=0#( as #0/x=0#) Transfer the #5# #x=5# Voila! Shreya May 14, 2018 Answer: see below Explanation: We have, #x^2-5x=0# #x(x-5)=0# Thus,either #x# can be #0# or it can be #5# 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 197 views around the world You can reuse this answer Creative Commons License