Quadratic Inequalities

1. One side is already zero. Find the roots by factorising.
   $$(x+2)(x-7)=0$$
   $$x=-2 \quad\text{or}\quad x=7$$
2. Sketch it. Here $a=1>0$, so the curve is a U-shape cutting the axis at $-2$ and $7$.
3. You want the parts above the axis. A U-shape is above the axis outside its roots.

1. Bring everything to one side first.
   $$2x^2+3x-27\le0$$
2. Find the roots.
   $$(2x+9)(x-3)=0$$
   $$x=-\tfrac{9}{2} \quad\text{or}\quad x=3$$
3. This curve is also a U-shape, and you want the part on or below the axis. That is the piece between the roots.

The roots are $-2$ and $7$. Since the coefficient of $x^2$ is negative, the curve opens down and is above the axis between its roots.

Since $(x+2)^2\ge0$, the expression is at least $1$. Therefore $x^2+4x+5>0$ for every real value of $x$, and $x^2+4x+5<0$ has no real solution.

$(x-3)^2<0$: no real solution.