Help with find the zero's of natural log function

Find all real zeros of f:

f(x)= ln(3x^2 + 27)

this has no real roots but 2 complex roots
ln(3x^2+27 = 0
3x^2+27 = e^0 = 1
x^2 = -26/3
x = ±i*sqrt(26/3)

This logarithmic can never be zero with real roots, it only has imaginary roots.

ln1 = 0

So when does 3x^2 + 27 = 1?

Let's see:
3x^2 = -26
x^2 = -26/3
x = sqrt(-26/3)

This has no real solutions since you cannot have a negative square root.