(x^3 - 3x^2 - 5x + 39) = (x^2 - 6x + 13)(x + 3). [you can verify this by expanding the brackets]
x + 3 = 0 when x = -3
x^2 - 6x + 13 = 0
x^2 - 6x + 9 + 4 = 0
(x - 3)^2 = -4
This shows x^2 - 6x + 13 = 0 has no real roots
The function only has one real root at x = -3