How do you solve #(x+2) (2x+5)=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer James May 20, 2018 #x=-2# or #x=-5/2# Explanation: show below #(x+2) (2x+5)=0# #x+2=0rArr x=-2# or #2x+5=0 rArr 2x=-5 rArr x=-5/2# 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 1446 views around the world You can reuse this answer Creative Commons License