How do you solve #(4x+1)(2x-7) =0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Konstantinos Michailidis Sep 22, 2015 Refer to explanation Explanation: When you have #a*b=0# then #a=0# or #b=0# hence we have that #(4x+1)*(2x-7)=0=>4x+1=0 or 2x-7=0=>x=-1/4 or x=7/2# 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 3870 views around the world You can reuse this answer Creative Commons License