How do you solve #3(y-5)^2=-12#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nghi N Jul 8, 2017 2 imaginary roots: #y = 5 +- 2i# Explanation: #(y - 5)^2 = - 12/3 = - 4 = 4i^2# #y - 5 = +- 2i# There are 2 imaginary roots: #y = 5 +- 2i# 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 911 views around the world You can reuse this answer Creative Commons License