How do you solve #x^2-9=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Apr 28, 2016 #x=-3# or #x=3# Explanation: In #x^2-9=0#, add and subtract #3x#, this leads to #x^2+3x-3x-9=0# or #x(x+3)-3(x+3)=0# or #(x-3)(x+3)=0# As product of #x+3# and #x-3# is zero, either #x+3=0# or #x-3=0# i.e. #x=-3# or #x=3# 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 855 views around the world You can reuse this answer Creative Commons License