How do you solve the equation #-4(x^2-8)=16#? Algebra Quadratic Equations and Functions Use Square Roots to Solve Quadratic Equations 1 Answer EZ as pi May 21, 2017 #x = +2 or x=-2# Explanation: There is no #x#=term, so we can solve this by finding the square roots. #-4(x^2-8)=16" "larr div -4# #x^2 -8= -4" "larr# add 8 to both sides #x^2 = -4+8# #x^2 = 4# #x = +-sqrt4# #x =+-2# 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 1410 views around the world You can reuse this answer Creative Commons License