Answer
The function has a minimum value of $-21$.
Work Step by Step
RECALL:
(1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens:
(i) upward when $a \gt0$ and has its vertex as its minimum; or
(ii) downward when $a\lt0$ and has its vertex as its maximum.
(2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$
Compare $f(x)=2x^2+12x-3$ to $f(x)=ax^2+bx+c$.
It can be seen that $a=2$, $b=12$, $c=-3$.
$a\gt0$, hence the graph opens up, hence it's vertex is a minimum.
The minimum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot 2}=-3.$
Thus, the minimum value is $f(-3)=2\cdot(-3)^2+12\cdot(-3)-3=-21.$