How do you solve #x^2+8=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nam D. Mar 20, 2018 Given: #x^2+8=0# Subtract #8# from both sides. #x^2=0-8# #=-8# Take square root of both sides. #x=sqrt(-8)# #=sqrt(4*-2)# #=sqrt(4)*sqrt(-2)# #=2sqrt(-2)# #=2sqrt(2)*sqrt(-1)# #=2sqrt(2)i# #=2isqrt(2)# 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 4936 views around the world You can reuse this answer Creative Commons License