Help minimum value of natural log

What is the minimum value of: e^(t^2+5t+6)

f(t) = e^(t^2+5t+6)
f'(t) = (2t+5)e^(t^2+5t+6) = 0 for extrema
2t+5 = 0
t = -5/2
f(-5/2) = e^(-0.25) = 0.7788
Minimum value of the function = 0.7788

y = e^(t²+5t+6)
y' = e^(t²+5t+6)(2t+5) = 0
t = -5/2

y = e^[(-5/2)² + 5(-5/2)+6]
y = e^(25/4 - 25/2 + 6)
y = e^(6-25/4)
y = e^(-1/4)
y ≈0.77880

The minimum value of e^(t²+5t + 6) is e^(-1/4) ≈ 0.77880

the term natural log means log with base e .ln

i.e

ln e^(t^2+5t+6)

ln n^m=m ln n

(t^2+5t+6) ln e

ln e =1 as its natural log

so

answer is (t^2+5t+6)

t=0