Second-degree inequalities

The discriminant of the associated equation equals $25>0$, so the trinomial has the two roots $-2$ and $3$. The leading coefficient is $1>0$, therefore the parabola opens upward and the trinomial is negative between the roots, zero at the roots, positive outside.

The inequality asks for the points where the trinomial is negative or zero, so the solutions are the points of the interval between the roots, endpoints included: $-2\le x\le 3$.

The discriminant is zero, so the trinomial has only the double root $x_0=\frac{1}{2}$ and can be written as a square: $4x^2-4x+1=(2x-1{)}^2$. A square is positive at every point other than the root of its base and zero at that root.

The first inequality asks where the square is positive, therefore the solutions are all real numbers other than $\frac{1}{2}$. The second asks where the square is negative or zero; since a square is never negative, the only solution is the point $x=\frac{1}{2}$ where it vanishes.

In the first inequality the discriminant is negative and the leading coefficient is positive, so the parabola lies entirely above the $x$-axis and the trinomial is positive at every point. Every real number is therefore a solution.

The second inequality has a negative leading coefficient, so both sides are multiplied by $-1$ reversing the direction, and the inequality becomes $2x^2-3x+4\le 0$. The discriminant equals $-23<0$ and the leading coefficient is positive, therefore the trinomial $2x^2-3x+4$ is positive at every point and is never less than or equal to zero. Hence the inequality has no solutions.