Solve the following equations by using formula:

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4$

Given,

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4 \\[1em] \Rightarrow \dfrac{(x - 2)^2 + (x + 2)^2}{(x - 2)(x + 2)} = 4 \\[1em] \Rightarrow \dfrac{x^2 + 4 - 4x + x^2 + 4 + 4x}{x^2 - 2x + 2x - 4} = 4 \\[1em] \Rightarrow \dfrac{2x^2 + 8}{x^2 - 4} = 4 \\[1em] \Rightarrow 2x^2 + 8 = 4(x^2 - 4) \\[1em] \Rightarrow 2x^2 + 8 = 4x^2 - 16 \\[1em] \Rightarrow 2x^2 - 4x^2 + 8 + 16 = 0 \\[1em] \Rightarrow -2x^2 + 24 = 0 \\[1em] 2x^2 - 24 = 0 \text{ (Multiplying equation by -1) }$

The equation is $2x^2 - 24 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 0 , c = -24

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-0 ± \sqrt{0^2 - 4 \times 2 \times (-24)}}{2 \times 2} \\[1em] \Rightarrow \dfrac{ ±\sqrt{192}}{4} \\[1em] \Rightarrow \dfrac{±8\sqrt{3}}{4} \\[1em] +2\sqrt{3} \text { or } -2\sqrt{3} \\[1em]$

Hence roots of the given equations are $2\sqrt{3} , -2\sqrt{3}$.