Answer
See below.
Work Step by Step
Let's compare $f(x)=2x^2+12x-3$ to $f(x)=ax^2+bx+c$. We can see that a=2, b=12, c=-3. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot 2}=-3.$ Hence the minimum value is $f(-3)=2(-3)^2+12(-3)-3=-21.$
