Read it again any time, including while you work through the exercises.
You know how to solve ax² + bx + c = 0 with the discriminant. A new, richer question: for which values of x is the quadratic positive? Negative? This is the study of the SIGN of the quadratic, and it is the key to second-degree inequalities like x² - 5x + 6 > 0.
Keep this picture in mind: the graph of a quadratic is a parabola. If a > 0 it smiles (opens upward); if a < 0 it frowns (opens downward). The roots are where it crosses the horizontal axis. Between the roots, the parabola dips to the other side of the axis: there, the sign is opposite to the sign of a.
Hence the golden rule, when Δ > 0 and the quadratic has two roots x₁ < x₂: the quadratic has the sign of a OUTSIDE the roots, and the OPPOSITE sign of a between the roots. Example: x² - 5x + 6 has roots 2 and 3, with a = 1 > 0. So it is positive before 2 and after 3, negative between 2 and 3, and zero at 2 and at 3.
The two other cases can be read off the parabola. If Δ = 0, the parabola touches the axis at a single point: the quadratic has the sign of a everywhere except at the root, where it vanishes. If Δ < 0, the parabola never touches the axis: the quadratic keeps the sign of a for ALL x. Thus x² + x + 1, with Δ = -3, is always strictly positive.
To solve a quadratic inequality, the ritual is: bring everything to one side, compute Δ and the possible roots, draw the sign table (or picture the parabola), then read off the requested zone. One last point of care: a strict inequality (< or >) excludes the roots, a non-strict one (≤ or ≥) includes them.