How do you solve #(x-2)^2=-3#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Ratnaker Mehta Aug 11, 2016 #(1)# : In #RR#, the soln. set is #phi#. #(2)# : #In CC#, the soln set is #{2+-isqrt3 }.# Explanation: In #RR#, the soln. set is #phi#; as, in #RR, (x-3)^2>=0# In #CC# since, #i=sqrt(-1)#, we have, #(x-2)^2=-3=(isqrt3)^2# #rArr x-2=+-isqrt3 rArr x=2+-isqrt3# 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 912 views around the world You can reuse this answer Creative Commons License