2x^3 + 12x^2 - 72x + 4
Tangent line horizontal means slope = 0.
Slope is first derivative, so:
f(x) = 2x^3 + 12x^2 - 72x + 4
f ' (x) = 6x^2 + 24x - 72
6x^2 + 24x - 72 = 0 solve for x
6(x^2 + 4x - 72) = 0
x^2 + 4x - 72 = 0
Doesn't look like it factors easy, so let's try the quadratic formula.
x = (-4 +- sqrt(16 - 4(1)(-72) ) / 2
x = (-4 +- sqrt(16 + 288) ) / 2
x = (-4 +- sqrt(304) ) / 2
x = (-4 +- sqrt(16*19) ) / 2
x = (-4 +- 4sqrt(19) ) / 2
x = -2 +- 2sqrt(19)
The tangent line will be horizontal at x = -2 + 2sqrt(19) and x = -2 - 2sqrt(19).