How do you solve #(x + 4)^2 = 11#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. Oct 29, 2015 #x=-4-sqrt(11)color(white)("XXX")#or#color(white)("XXX")x=-4+sqrt(11)# Explanation: Given #color(white)("XXX")(x+4)^2=11# then #color(white)("XXX")x+4 = +-sqrt(11)# and #color(white)("XXX")x=-4+-sqrt(11)# 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 1834 views around the world You can reuse this answer Creative Commons License