How do you solve #x^2 = 2x + 15#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Apr 9, 2016 #x=-3# or #x=5# Explanation: #x^2=2x+15# #hArrx^2-2x-15=0# Splitting the middle term in #-5x# and #3x# we get #x^2-5x+3x-15=0# or #x(x-5)+3(x-5)=0# or #(x+3)(x-5)=0# Hence #x=-3# or #x=5# 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 941 views around the world You can reuse this answer Creative Commons License