Решите уравнение f’(t)=0, если f(t)= (2t+3)^2(t-3)

f'(t) = 2*(2t+3)*2*(t-3) + (2t+3)^2*1 = 4 (2t^2+3t-6t-9) + 4t^2+12t+9 = 8t^2-12t-36+4t^2+12t+9 = 12t^2-27t^2=27/12=9/4t= +- 3/2