Solution by Factorisation

If we can factorise $ax^2 + bx + c = 0$ as $(px + q)(rx + s) = 0$, then:

$$x = -\frac{q}{p} \quad \text{or} \quad x = -\frac{s}{r}$$

Example

Solve $x^2 - 5x + 6 = 0$.

$$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$ $$\therefore \; x = 2 \quad \text{or} \quad x = 3$$