What is the solution set for #x^2 - 2x + 5 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nallasivam V Aug 10, 2015 #x =(2+ sqrt( - 16))/ (2)# #x =(2- sqrt( - 16))/ (2)# Explanation: Since ------- #(-2)^2# - #(4 xx1 xx 5)# < 0, x has imaginary roots #x =(-b+- sqrt(b^2 - (4ac)))/ (2a)# #x =(-(-2)+- sqrt((-2)^2 - (4xx1xx5)))/ (2xx1)# #x =(-(-2)+- sqrt(4 - 20))/ (2)# #x =(2+ sqrt( - 16))/ (2)# #x =(2- sqrt( - 16))/ (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 4968 views around the world You can reuse this answer Creative Commons License