One negative real around -0.8 and two complex conjugate zeros.

https://www.wolframalpha.com/input/?i=f(x)+%3D+6x%...

f(x) = 6x^3 − 3x^2 + 5x + 9

so

by Descartes rule of signs , there are two sign changes in original equation

so there are 2 or 0 positive real roots

replacing x with (-x)

f(-x) = 6(-x)^3 -3(-x)^2 +5(-x) + 9

f(-x) = -6x^3 -3x^2 -5x + 9

so one sign change , 1 negative real root

since there 3 roots totals

you have either 2 or 0 positive real roots

you have 2 or 0 complex roots (there only come in pairs)

you have 1 negative real root

f'(x) = 18x^2 - 6x + 5,

which is never zero

(because the "discriminant" is negative).

Therefore, the function increases monotonically, and has exactly one real zero.

Because f(0) = 9, the real zero of f(x) must be in the negative domain. Looks like it's pretty close to -2/3, since I note f(-1) = -5.

There must be two other zeros, though (because it's a cubic function). The other two zeros are complex.

There are three zeros for f(x), 1 negative and 2 complex

For any polynomial function, the list of possible rational roots can be made by looking at the factors of the constant term over the factors of the high-degree coefficient.

Factors of 9: ±1, ±3, ±9

Factors of 6: ±1, ±2, ±3, ±6

Putting all combinations together we get:

±1, ±3, ±9, ±1/2, ±3/2, ±9/2, ±1/3, ±1/6

(note that 3/3, 9/3, 3/6, and 9/6 aren't included since they reduce down to a number that we already have included, so I didn't include them as duplicates).

So if you were to use this list to brute-force your options to see if any are zeroes, you have 16 options to test.

And none of them would be zeroes.

There is one irrational root and two complex roots.