How do you solve #x^2 + 16x + 64 = 81#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer sjc Mar 21, 2018 #x=1, -17# Explanation: rearrange to the form #ax^2+bx+c=0# then factorise #x^2+16x+64=81# #x^2+16x-17=0# #(x+17)(x-1)=0# #" either "x+17=0=>x=-17# #" or "x-1=0=>x=1# #:.x=1,-17# 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 1873 views around the world You can reuse this answer Creative Commons License