$f(x)=ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}≥c-\frac{b^2}{4a}$ if $a>0$ as $(x+\frac{b}{2a})^2≥0$ for real $x,a,b$.
So, the minimum value of $f(x)=ax^2+bx+c$ is $c-\frac{b^2}{4a}$ if $a>0$
Putting $a=3,b=12,c=180$, $c-\frac{b^2}{4a}=180-\frac{(12)^2}{4\cdot 3}$ $=180-12=168=168.00$ (with 2 decimal places precision)
The extreme value of $ax^2+bx+c$ can be calculated using another approach apart from differentiation as follows:
Let $y=ax^2+bx+c\implies ax^2+bx+c-y=0$
As $x$ is real, $(-b)^2≥4\cdot a\cdot(c-y)\implies \frac{b^2}{4a}≥c-y\implies y≥c-\frac{b^2}{4a} $
So, $ax^2+bx+c=y≥c-\frac{b^2}{4a}$