Answer: \(x=3\sqrt5-5\), \(y=20\sqrt5\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

Differentiate the curve:

\(\frac{dy}{dx}=2(3+\sqrt5)x-8\sqrt5.\)

At a stationary point, \(\frac{dy}{dx}=0\), so

\(2(3+\sqrt5)x-8\sqrt5=0.\)

Therefore

\(x=\frac{4\sqrt5}{3+\sqrt5}.\)

Rationalise the denominator:

Thus

\(x=3\sqrt5-5.\)

Now substitute \(x=3\sqrt5-5\) into the curve.

First,

\((3\sqrt5-5)^2=70-30\sqrt5.\)

Hence

\(y=(3+\sqrt5)(70-30\sqrt5)-8\sqrt5(3\sqrt5-5)+60.\)

Simplifying,

\((3+\sqrt5)(70-30\sqrt5)=60-20\sqrt5\)

and

\(-8\sqrt5(3\sqrt5-5)=-120+40\sqrt5.\)

Therefore

\(y=(60-20\sqrt5)+(-120+40\sqrt5)+60 =20\sqrt5.\)