Solve $\frac{1}{x}-\frac{1}{x+4}=\frac{1}{3}$, where $x$ ≠ 0 and $x$ ≠ −4.

$D=b^2-4ac$ classifies the roots of $a{x}^2+bx+c=0:-\ast \ast D>0$**: TWO distinct real roots. Parabola crosses $x$-axis at two points.
- **$D=0$**: ONE repeated real root (called a 'double root' or 'two equal roots'). Parabola touches $x$-axis at the vertex.
- **$D<0$**: NO real roots. Parabola does not touch the $x$-axis at all.

**Use case**: 'For what values of $k$ does $k{x}^2+2x+1=0$ have equal roots?' → set $D=0:4-4k=0$ → $k=1$.

You can answer 'how many real roots?' WITHOUT computing them — just compute $D$.