Help with maths?

9x^2+12x+k = 0
gives no real value
b^2<4ac
144 < 36k
k > 4

y = 3x³ + 6x² + kx + 6 ← this is a curve (a function)

y' = 9x² + 12x + k ← this is the derivative

...and you can get a stationary point when the derivative is null.

No stationary point means that the derivative is never null. You have to solve: y' ≠ 0

9x² + 12x + k ≠ 0

Polynomial like: ax² + bx + c, where:
a = 9
b = 12
c = k

Δ = b² - 4ac (discriminant)

Δ = 144 - 36k → no solution for the equation → no real roots → Δ < 0

144 - 36k < 0

36.(4 - k) < 0

4 - k < 0

- k < - 4

k > 4

dy/dx = 9x² + 12x + k = 0 at stationary points
9x² + 12x + k ≠ 0 if discriminant < 0
discriminant = 12² - 4·9·k = 144-36k < 0
k > 4

y=3x^3+6x^2+kx+6
y' = 9x^2 + 12x + k

y'>0 for all x when k>4,
so the graph y=3x^3+6x^2+kx+6 will have no stationary points when k>4