How do you solve #Y = (x - 5)² - 9#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Adrian D. Jul 20, 2015 #x=5+-sqrt(Y+9)# Explanation: You have the solution of #Y# in terms of #x#, to reverse this and solve for #x# in terms of #Y# we have: #Y=(x-5)^2-9# => #(x-5)^2=Y+9# => #x=5+-sqrt(Y+9)# 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 2641 views around the world You can reuse this answer Creative Commons License