If $${\text{2}}x - \frac{1}{{2x}} = 5{\text{,}}$$    $${\text{x}} \ne {\text{0,}}$$   then find the value of $${x^2} + \frac{1}{{16{x^2}}} - 2$$    = ?

$$\eqalign{ & 2x - \frac{1}{{2x}} = 5 \cr & {\text{Divide by 2 both side}} \cr & x - \frac{1}{{4x}} = \frac{5}{2} \cr & {\text{Squaring both side}} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr & {\text{So, }} \cr & {x^2} + \frac{1}{{16{x^2}}} - 2 \cr & = \frac{{27}}{4} - 2 \cr & = \frac{{19}}{4} \cr} $$