Answer: \(x=\dfrac6{29}+\dfrac{\sqrt7}{29}\).

For

\((5+4\sqrt7)x^2+(4-2\sqrt7)x-1=0,\)

use the quadratic formula with

\(a=5+4\sqrt7,\qquad b=4-2\sqrt7,\qquad c=-1.\)

The positive solution is

\(x=\frac{-(4-2\sqrt7)+\sqrt{(4-2\sqrt7)^2-4(5+4\sqrt7)(-1)}}{2(5+4\sqrt7)}.\)

Simplify the discriminant:

\((4-2\sqrt7)^2-4(5+4\sqrt7)(-1) =16+28-16\sqrt7+20+16\sqrt7=64.\)

So

\(x=\frac{-(4-2\sqrt7)+8}{2(5+4\sqrt7)} =\frac{4+2\sqrt7}{2(5+4\sqrt7)}.\)

Hence

\(x=\frac{2+\sqrt7}{5+4\sqrt7}.\)

Rationalise the denominator:

\(x=\frac{2+\sqrt7}{5+4\sqrt7}\cdot\frac{5-4\sqrt7}{5-4\sqrt7}.\)

The numerator is

\(10-8\sqrt7+5\sqrt7-28=-18-3\sqrt7,\)

and the denominator is

\(25-16(7)=-87.\)

Therefore

\(x=\frac{18+3\sqrt7}{87} =\frac6{29}+\frac{\sqrt7}{29}.\)