How do you solve #3/4(2x+3)^2-9=0 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer P dilip_k Apr 30, 2016 #x=(-3+-2sqrt3)/2# Explanation: #3/4(2x+3)^2-9=0 # #=>3/4(2x+3)^2=9 # #=>(2x+3)^2=9xx4/3=12 # #=>(2x+3)=+-sqrt(3*2^2) =+-2sqrt3# #=>2x=(-3+-2sqrt3)# #=>:.x=(-3+-2sqrt3)/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 1340 views around the world You can reuse this answer Creative Commons License