How do you find all the complex roots of #x^2 + 8x + 21=0#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer Konstantinos Michailidis Jan 22, 2016 Rewrite this as follows #x^2+8x+16+5=0=>(x+4)^2=-5=>(x+4)^2=5i^2=> x_1=-4+sqrt5*i , x_2=-4-sqrt5*i# where #i# is the imaginary unit with #i^2=-1# 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 1603 views around the world You can reuse this answer Creative Commons License