Explanation

(c)
Given, $\sin{\pi\left( x^{2} + x \right) - \sin{\pi x^{2} =}0}$
$\Rightarrow \ 2\cos{\pi\left( \frac{2x^{2} + x}{2} \right)\sin{\frac{\pi x}{2} = 0}}$
$\Rightarrow \ \pi\left( \frac{2x^{2} + x}{2} \right) = n\pi + \frac{\pi}{2}$
$\Rightarrow \ \ 2x^{2} + x = 2n + 1$
$\Rightarrow \ \ 2x^{2} + x - p^{'} = 0$, where $p^{'} = 2n + 1$, is an odd integer
$\therefore\ \ x = \frac{- 1 \pm \sqrt{1 + 8p'}}{4}$ [put $1 + 8p^{'} = p$]
$\therefore\ \ x = \frac{- 1 \pm \sqrt{p}}{4}\ \ \Rightarrow \ \ x = \frac{\sqrt{p} - 1}{4}\ \ \ \ \ \ \ \left\lbrack neglect\ x = \frac{- 1 - \sqrt{p}}{4} \right\rbrack$