How do you solve #2x ^ { 2} + 10x - 168= 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer sankarankalyanam Oct 16, 2017 #**x = -12, 7**# Explanation: #2x^2+10x-168=0#. Dividing by 2, #x^2+5x-84=0# #x^2+12x-7x-84=0# #x(x+12)-7(x+12)=0# #(x+12)(x-7)=0# #x = -12, 7# 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 1716 views around the world You can reuse this answer Creative Commons License