How do you solve # x²-48=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Kevin B. Jun 7, 2015 #x = +-4sqrt(3)# Explanation: #x^2 - 48 = 0# #x^2 = 48# #x = +-sqrt(48)# #x = +-sqrt(16 *3)# #x = +-4sqrt(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 5447 views around the world You can reuse this answer Creative Commons License