Alright so you have \(x^{2} -18x+8\)
You would think to factor this that you need to find two numbers that add up to the coefficient of the middle term (-18) and multiply to the last term (8).
But this is not factorable
So to solve the equation \(x^2-18x +8 = 0\) you have to use the quadratic formula, which is \(x = (-b + \sqrt{b^2-4ac}) / 2a, (-b - \sqrt{b^2-4ac}) / 2a\)
This polynomial is already in standard form \(ax^2+bx+c\), where a is -18, b is -18, and b is 8, so just plug the numbers into the equation.
So you get \(x = 9 + \sqrt{73}\), \(x=9- \sqrt{73}\)
