How do you solve #(x – 6)^2 = 2#? Algebra Quadratic Equations and Functions Use Square Roots to Solve Quadratic Equations 1 Answer EZ as pi Jul 16, 2016 #x =7.414 or 4.586# Explanation: This is in the form #x^2 = y#, so we can just find the square root of both sides without having to multiply out first. #(x - 6)^2 = 2# #x - 6= +-sqrt2# #x = +sqrt2 +6 = 7.414# #x = -sqrt2 +6 = 4.586# Answer link Related questions Can you use square roots to solve all quadratic equations? How do you use square roots to solve quadratic equations? How do you solve #4x^2-49=0# by taking square roots? How do you solve #(x-3)^2+25=0#? How do you use square roots to solve #2(x+3)^2=8#? What is Newton's formula for projectile motion? How do you solve for x in the equation #(x-3)^2+25=0#? How do you solve the equation #3(x-2)^2-12=0# How do you solve #3x^2+8x=9+2x#? How do you solve #4x^2=52# solve using the square root property? See all questions in Use Square Roots to Solve Quadratic Equations Impact of this question 1435 views around the world You can reuse this answer Creative Commons License