Explanation

(d)
Given, $x^{3} + 6x + 9 = 0$
$\Rightarrow \ \ \ (x + 3)\left( x^{2} - 3x + 3 \right) = 0$
$\Rightarrow x = - 3$ or $x^{2} - 3x + 3 = 0$
Now, Discriminant, $D = \sqrt{9 - 4 \times 3} = \sqrt{- 3}$ imaginary
Hence, real roots of the given equation is $- 3$