How do you solve # n^2 + 3n - 12 = 6#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nam D. Apr 29, 2018 #n=-6,3# Explanation: Given: #n^2+3n-12=6#. Subtract #6# from both sides. #n^2+3n-12-6=color(red)cancelcolor(black)6-color(red)cancelcolor(black)6# #n^2+3n-18=0# #=>(n+6)(n-3)=0# #:.n=-6,3# 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 4406 views around the world You can reuse this answer Creative Commons License