How do you solve #x^2 + x – 12 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer ali ergin Mar 30, 2016 #x={3,-4}# Explanation: #x^2+x-12=0# #(x+4)*(x-3)=0# #if (x+4)=0" "then" "x=-4# #if (x-3)=0" "then" "x=3 # #x={3,-4}# 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 1027 views around the world You can reuse this answer Creative Commons License