Answer: \(x=1+\frac12\sqrt{10}\) or \(x=-\frac23-\frac13\sqrt{10}\).

We start with the main method. Use the quadratic formula exactly and simplify the surds by rationalising where necessary.

For

\((2-\sqrt{10})x^2+x+(2+\sqrt{10})=0,\)

use the quadratic formula with

\(a=2-\sqrt{10},\quad b=1,\quad c=2+\sqrt{10}.\)

The discriminant is

\(b^2-4ac=1-4(2-\sqrt{10})(2+\sqrt{10}).\)

Now

\((2-\sqrt{10})(2+\sqrt{10})=4-10=-6.\)

So

\(b^2-4ac=1-4(-6)=25.\)

Therefore

\(x=\frac{-1\pm5}{2(2-\sqrt{10})}.\)

Using the positive sign gives

\(x=\frac4{4-2\sqrt{10}}=\frac2{2-\sqrt{10}}.\)

Rationalising,

\(x=\frac{2(2+\sqrt{10})}{4-10}=-\frac{2+\sqrt{10}}3.\)

So one solution is

\(x=-\frac23-\frac13\sqrt{10}.\)

Using the negative sign gives

\(x=\frac{-6}{4-2\sqrt{10}}=-\frac3{2-\sqrt{10}}.\)

Rationalising,

\(x=-\frac{3(2+\sqrt{10})}{4-10}=\frac{2+\sqrt{10}}2.\)

So the other solution is

\(x=1+\frac12\sqrt{10}.\)

This completes the solution and gives the required result.