How do you solve #(x-5)(x+6)=12#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Evan Apr 28, 2018 #x=6 or -7# Explanation: #(x-5)(x+6)=12# Expand, #x^2+x-30=12# Subtract #12# from both sides, #x^2+x-42=0# Factor, #(x-6)(x+7)=0# Solve, #x=6 or -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 2895 views around the world You can reuse this answer Creative Commons License