How do you factor and solve #25x^2 -9 =0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer sente Jul 12, 2016 #x in {-3/5, 3/5}# Explanation: Using the difference of squares formula #a^2-b^2 = (a+b)(a-b)#, we have #25x^2 - 9 = (5x)^2-3^2=(5x+3)(5x-3) = 0# #=>5x+3 = 0 or 5x-3 = 0# #=> 5x = -3 or 5x = 3# #=> x = -3/5 or x = 3/5# 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 5691 views around the world You can reuse this answer Creative Commons License