I assume you actually meant :

f(t) = t^4 +t^3 - 18t^2 +4

if so

f'(t) = 4t^3 +3t^2 -36t +0

and f"(t) = 12t^2 +6t -36

or

f"t) = 6(2t^2 +t -18)

now, we set this = 0 and solve for the two x's , which will be the inflection points

It does not factor, so we MUST use the quadratic formula

t = (-1±√145)/4 <----- exact answers

or

Numerical approximation answers :

t1 = 2.76034 ( Larger answer inflection point )

t2 = -3.26039...( Smaller inflection point )

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Confirm :

Plot the original quartic,

y1 = x^4 +x^3 - 18x^2 +4 carefully, on a TI-84, and check for the two IP