How do you find the zeroes of #g(x)=2x^2-5x-3 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. Jun 7, 2015 #g(x) = 2x^2 -5x-3#can be factored as #color(white)("XXXX")##g(x)=(2x+1)(x-3)# The zeroes of #g(x)# are the values of #x# for which #g(x)=0# If #g(x) = (2x+1)(x-3) = 0# then either #color(white)("XXXX")##2x+1 = 0 rArr x= -1/2# or #color(white)("XXXX")##x-3 = 0 rArr x = 3# So the zeros of #g(x)# are #(-1/2)# and #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 2256 views around the world You can reuse this answer Creative Commons License