Explanation

(c)
Let $y = 2x^{2} + x - 1$
$y = 4x + 1$
For maxima or minima, put $y^{'} = 0$
$\Rightarrow \ \ \ \ \ \ x = - \frac{1}{4}$
Now, $y^{''} = 4 = +$ ve
$y$ is minimum at $x = - \frac{1}{4}$
Thus, minimum value $= 2\left( - \frac{1}{4} \right)^{2} + \left( - \frac{1}{4} \right) - 1 = - \frac{9}{8}$
Alternate
Here $a > 0$
$\therefore$ Minimum value$= \frac{4ac - b^{2}}{4a}$
$= \frac{4 \times 2( - 1) - 1}{4 \times 2} = - \frac{9}{8}$