Answer: D

Small Hint:

Complete the square in $ax^2+bx+c$

Big Hint:

Since $a>0,$ the squared term is minimized when it equals zero

Solution:

Completing the square gives $$\begin{aligned} ax^2+bx+c &=a\left(x+\frac{b}{2a}\right)^2\\ &\quad+c-\frac{b^2}{4a}. \end{aligned}$$ Because $a>0,$ the squared term has minimum $0.$ The least value is therefore $$c-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}.$$

Thus, the correct answer is D.