please help me with this. I don't understand what my professor is asking....

my original question and answer:

15t²+ 2t+5 = 3t(t²+1). Let's call this is equation (1)

We'll have: (1) 15t²+ 2t+ 5 = 3t³+ 3t

15t²+ 2t+ 5 - 3t³- 3t = 0

-3t³+ 15t²- t + 5 = 0

(t-5) (-3t²-1) = 0

t -5 = 0 or -3t²-1 =0

t=5 or 3t²+1 = 0 (divide by -1)

Consider 3t²+ 1. We have t²always greater or equal than 0, so 3t²must be greater or equal than 0. Therefore, 3t²+1 must be greater than 0. Therefore, there is no value of t for the equation3t²+ 1.

Overall, there is only one value of t =5 that corrects to the (1) equation.

Her reply:

You are correct that there is only one real number solution of the equation, and that is t = 5. However, the exercise asks for all complex solutions, not just real number solutions.

So please complete the problem by also solving the equation 3t²+ 1 = 0 for the complex number solutions. Thanks!