Answer: \(x=-6+3\sqrt5\) or \(x=-2+\sqrt5\).

Rewrite the equation as a standard quadratic in the chosen variable, then solve and reject any value that does not satisfy the original form.

Rewrite the equation as

\((2+\sqrt5)x^2-4x-3(2-\sqrt5)=0.\)

Using the quadratic formula,

\(x=\frac{4\pm\sqrt{16+12(2+\sqrt5)(2-\sqrt5)}}{2(2+\sqrt5)}.\)

Since

\((2+\sqrt5)(2-\sqrt5)=4-5=-1,\)

this becomes

\(x=\frac{4\pm\sqrt{16-12}}{2(2+\sqrt5)} =\frac{4\pm2}{2(2+\sqrt5)}.\)

So

Rationalising,

\(\frac{3}{2+\sqrt5} =\frac{3(2-\sqrt5)}{4-5} =-6+3\sqrt5,\)

and

\(\frac{1}{2+\sqrt5} =\frac{2-\sqrt5}{4-5} =-2+\sqrt5.\)

Therefore the result matches the required answer.