How do you solve #(x + 6)^2 = 121#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. May 8, 2016 #x=-17# or #x=5# Explanation: Given #color(white)("XXX")(x+6)^2=121# Therefore #color(white)("XXX")x+6 = +-sqrt(121) = +- 11# If #color(white)("XXX"){: (x+6=-11,color(white)("XXXX"),x+6=+11), (rarrx=-17,,rarrx=+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 2398 views around the world You can reuse this answer Creative Commons License