How do you solve #2x^2+12=0#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer Shaikh A. Dec 6, 2016 #x=+-sqrt6 i# Explanation: #2x^2+12=0# #2(x^2+6)=0# #x^2=-6# Taking square root both sides #x=+-sqrt6i# Answer link Related questions What is a complex conjugate? How do I find a complex conjugate? What is the conjugate zeros theorem? How do I use the conjugate zeros theorem? What is the conjugate pair theorem? How do I find the complex conjugate of #10+6i#? How do I find the complex conjugate of #14+12i#? What is the complex conjugate for the number #7-3i#? What is the complex conjugate of #3i+4#? What is the complex conjugate of #a-bi#? See all questions in Complex Conjugate Zeros Impact of this question 2718 views around the world You can reuse this answer Creative Commons License