Step-by-step explanation:

First, we'll solve the equation \(20x - \frac{1}{x} = 6\) for \(x\). Let's simplify the equation:

\[20x^2 - 1 = 6x\]

\[20x^2 - 6x - 1 = 0\]

Using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where \(a = 20\), \(b = -6\), and \(c = -1\).

\[x = \frac{6 \pm \sqrt{(-6)^2 - 4(20)(-1)}}{2(20)}\]

\[x = \frac{6 \pm \sqrt{36 + 80}}{40}\]

\[x = \frac{6 \pm \sqrt{116}}{40}\]

\[x = \frac{6 \pm 2\sqrt{29}}{40}\]

Now that we have the value of \(x\), we can calculate \(x^2 + \frac{1}{16x^2}\):

\[x^2 + \frac{1}{16x^2} = \left(\frac{6 \pm 2\sqrt{29}}{40}\right)^2 + \frac{1}{16\left(\frac{6 \pm 2\sqrt{29}}{40}\right)^2}\]

\[x^2 + \frac{1}{16x^2} = \frac{36 \pm 24\sqrt{29} + 4 \cdot 29}{1600} + \frac{40^2}{16(36 \pm 24\sqrt{29})^2}\]

After simplification, we get the value of \(x^2 + \frac{1}{16x^2}\) in terms of \(29\) as:

\[x^2 + \frac{1}{16x^2} = \frac{145 \pm 96\sqrt{29}}{1600} + \frac{1600}{16(36 \pm 24\sqrt{29})^2}\]