$f(x) = 2x^{2}-8x-3$
$f'(x) = 4x-8 $
For stationary point$:f'(x) = 0\implies 4x - 8 = 0 \implies x = 2$
Therefore, critical points $x = 0,2,5$
Now, $f''(x) = 4>0\:\: \textbf{(Minima)}$
For getting the minimum (or) least value , we should check all the value of critical points (stationary point and closed interval points).
For $x = 0: f(x) = -3$
For $x = 2: f(x) = 8 - 16 - 3 = -11$
For $x = 5: f(x) = 50 - 40 - 3 = 7 $
$\therefore x= 2 ,\text{minimum value of}\: f(x) = -11$
So, the correct answer is $(C).$