How do you solve #2x^2+x-3=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Anjali G Nov 27, 2016 #x=1,x=-3/2# Explanation: #2x^2+x-3=0# Solve by factoring: #2x^2-2x+3x-3=0# #2x(x-1)+3(x-1)=0# #(2x+3)(x-1)=0# #2x+3=0# #2x=-3# #x=-3/2# #x-1=0# #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 27983 views around the world You can reuse this answer Creative Commons License