Answer: \(x\lt3\) or \(x\gt15\).

Expand the left-hand side:

\((x-8)(x-10)=x^2-18x+80.\)

So

\(x^2-18x+80\gt35.\)

Hence

\(x^2-18x+45\gt0.\)

Factorise:

\((x-15)(x-3)\gt0.\)

The quadratic is positive outside its two roots, so

\(x\lt3\quad\text{or}\quad x\gt15.\)