Let \(y = \frac{1}{x^2}\). Then \(y^2 = \frac{1}{x^4}\).

Substitute into the equation:

\(18y^2 + y = 4\)

Rearrange to form a quadratic equation:

\(18y^2 + y - 4 = 0\)

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 18\), \(b = 1\), \(c = -4\):

\(y = \frac{-1 \pm \sqrt{1^2 - 4 \times 18 \times (-4)}}{2 \times 18}\)

\(y = \frac{-1 \pm \sqrt{1 + 288}}{36}\)

\(y = \frac{-1 \pm \sqrt{289}}{36}\)

\(y = \frac{-1 \pm 17}{36}\)

\(y = \frac{16}{36} \text{ or } y = \frac{-18}{36}\)

\(y = \frac{4}{9} \text{ or } y = -\frac{1}{2}\)

Since \(y = \frac{1}{x^2}\), solve for \(x\):

For \(y = \frac{4}{9}\):

\(\frac{1}{x^2} = \frac{4}{9}\)

\(x^2 = \frac{9}{4}\)

\(x = \pm \frac{3}{2}\)

For \(y = -\frac{1}{2}\), \(x^2\) cannot be negative, so ignore.

Thus, the real roots are \(x = 1.5\) and \(x = -1.5\).