How do you solve #- 16x^2 + 36 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Don't Memorise Jun 15, 2015 # color(blue)(x = +-3/2# Explanation: #- 16x^2 + 36 = 0# # 16x^2 - 36 = 0# Taking out a common term: #4(4x^2 -9) = 0# # 4x^2 -9= 0# # x^2 = 9/4# # x = sqrt(9/4)# # color(blue)(x = +-3/2# 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 3328 views around the world You can reuse this answer Creative Commons License