How do you solve #t^2 - t = 12 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Deepak G. Jul 31, 2016 #t=4# #t=-3# Explanation: #t^2-t=12# or #t^2-t-12=0# or #t^2-4t+3t-12=0# or #t(t-4)+3(t-4)=0# or #(t-4)(t+3)=0# or #(t-4)=0# or #t=4#=======Ans #1# or #(t+3)=0# or #t=-3#=======Ans #2# 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 2315 views around the world You can reuse this answer Creative Commons License