How do you solve #x^2 - 5x - 6 = 0# by factoring? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer MeneerNask Aug 6, 2015 Since the #6# has a #-#sign, we're looking for two factors of #6# that have a difference of #5# (and thus opposite signs). Explanation: #6and 1# will do, where the #6# will be negative and the #1# positive, to get to the #-5# The factoring will be: #(x-6)(x+1)=0->x=6orx=-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 1019 views around the world You can reuse this answer Creative Commons License