(a)
Let $f(x) = - 3 + x - x^{2}$
Now, $D = 1^{2} - 4(3) = - 11 < 0$
Here, coefficient of $x^{2} < 0$
$\therefore\ \ f(x) < 0$
Thus, LHS of the given equation is always positive whereas the RHS is always less than zero
Hence, the given equation has no solution