Case a negative, Delta negative — always negative

With downward concavity and no zero ($\Delta<0$), the parabola lies entirely below the $x$-axis: it is always negative. An inequality asking for values $\ge 0$ then has no solution at all.

Example — Case $a < 0$, $\Delta < 0$, parabola always negative

Solve $-x^2+x-1 \ge 0$.

$\Delta = 1-4 = -3 < 0$ and $a=-1<0$: the parabola lies always below the $x$-axis.

The parabola lies entirely below the $x$-axis: it is negative for every value of $x$.

The inequality $-x^2+x-1 \ge 0$ asks where the parabola lies above or on the $x$-axis: never. Solution: $\boxed{\emptyset}$.

Links

Topics: Inequalities
Concepts: Discriminant · Second-degree inequality · Parabola
Skills: Reasoning by cases · Solving inequalities