How do you solve #k^ { 2} + 23k + 12= - 120#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Hamza The Syrian Mar 27, 2018 #k=-11# #or# #k=-12# Explanation: #k^2+23k+12=-120# #k^2+23k+12+120=0# #k^2+23k+132=0# #(k+11)(k+12)=0# #k+11=0# ==> #k=-11# #or# #k+12=0# ==> #k=-12# 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 1120 views around the world You can reuse this answer Creative Commons License