How do you solve the quadratic #9n^2-5=-99# using any method? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Alan N. Nov 25, 2016 #n=+-(sqrt94 i)/3# Explanation: #9n^2-5=-99# #9n^2=-94# #n^2=-94/9# #n=+-sqrt(-94/9)# #n=+-(sqrt94 i)/3# Answer link Related questions What does completing the square mean? How do I complete the square? Does completing the square always work? Is completing the square always the best method? Do I need to complete the square if #f(x) = x^2 - 6x + 9#? How do I complete the square if #f(x) = x^2 + 4x - 9#? How do I complete the square if the coefficient of #x^2# is not 1? How do I complete the square if #f(x) = 3x^2 + 12x - 9#? If I know the quadratic formula, why must I also know how to complete the square? How do I use completing the square to describe the graph of #f(x)=30-12x-x^2#? See all questions in Completing the Square Impact of this question 1319 views around the world You can reuse this answer Creative Commons License