Solve the following equation using the formula :

x² - 6x = 27

Given,

x² - 6x = 27

⇒ x² - 6x - 27 = 0

Comparing x² - 6x - 27 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -6, c = -27.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get,

$\Rightarrow x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-27)}}{2(1)} \\[1em] = \dfrac{6 \pm \sqrt{36 + 108}}{2} \\[1em] = \dfrac{6 \pm \sqrt{144}}{2} \\[1em] = \dfrac{6 + \sqrt{144}}{2} \text{ or } \dfrac{6 - \sqrt{144}}{2} \\[1em] = \dfrac{6 + 12}{2} \text{ or } \dfrac{6 - 12}{2} \\[1em] = \dfrac{18}{2} \text{ or } -\dfrac{6}{2} \\[1em] = 9 \text{ or } -3.$

Hence, x = -3, 9.