How do you solve #(x-2/3)^2 - 8= 1/3 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Cem Sentin Nov 28, 2017 #x_1=(2-5sqrt3)/3# and #x_2=(2+5sqrt3)/3# Explanation: #(x-2/3)^2-8=1/3# #(x-2/3)^2-25/3=0# #(x-2/3)^2-75/9=0# #(x-2/3)^2-((5sqrt3)/3)^2=0# #(x-2/3+(5sqrt3)/3)*(x-2/3-(5sqrt3)/3)=0# Hence #x_1=(2-5sqrt3)/3# and #x_2=(2+5sqrt3)/3# 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 961 views around the world You can reuse this answer Creative Commons License