How do you factor and solve #2x^2 + x - 1=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Suren Abreu Jan 15, 2017 #x=1/2# or #x=-1# Explanation: #2x^2+x-1=0# Factorise. #2x^2+2x-x-1=0# #2x(x+1)-1(x+1)=0# #(2x-1)(x+1)=0# #2x-1=0# or #x+1=0# #x=1/2# or #x=-1# 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 61217 views around the world You can reuse this answer Creative Commons License