How do you solve #5x^2 + 12r = -4x^2 -4#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. Jun 8, 2015 #5x^2+12r = -4x^2-4# #rArr 9x^2 = -12r-4# #rArr (3x)^2 = -12r -4# #rArr 3x = +-sqrt(-12r-4)# #rArr x = +- sqrt(-12r-4)/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 1129 views around the world You can reuse this answer Creative Commons License