Solve the following equation using the formula :

x² - 6 = $2\sqrt{2}$x

Given,

$\Rightarrow x^2 - 6 = 2\sqrt{2}x \\[1em] \Rightarrow x^2 - 2\sqrt{2}x - 6 = 0$

Comparing x² - $2\sqrt{2}$x - 6 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -$2\sqrt{2}$, c = -6.

We know that,

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

Substituting values we get,

$\Rightarrow x = \dfrac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2 - 4(1)(-6)}}{2(1)} \\[1em] = \dfrac{2\sqrt{2} \pm \sqrt{8 + 24}}{2} \\[1em] = \dfrac{2\sqrt{2} \pm \sqrt{32}}{2} \\[1em] = \dfrac{2\sqrt{2} \pm 4\sqrt{2}}{2} \\[1em] = \dfrac{2\sqrt{2} + 4\sqrt{2}}{2} \text{ or } \dfrac{2\sqrt{2} - 4\sqrt{2}}{2} \\[1em] = \dfrac{6\sqrt{2}}{2} \text{ or } \dfrac{-2\sqrt{2}}{2} \\[1em] = 3\sqrt{2} \text{ or } -\sqrt{2} \\[1em] = 4.23 \text{ or } -1.41$

Hence, x = 4.23, -1.41.