\(x= \frac{-b\pm \sqrt{b^2-4ac} }{2a}\) for the equation \(ax^2+bx+c\)
Your question is \(x^2-2x-8=0\)
So \(a=1 \\b=-2\\c=-8\)
So plug in your numbers:
\(x= \frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-8)} }{2(1)}\)
Simplify:
\(x= \frac{2\pm \sqrt{4-(-32)} }{2}\)
Simplify a little more:
\(\frac{2\pm6}{2}\)
So our answer is
\(x= \frac{2+ 6 }{2} = 4\\x= \frac{2- 6 }{2}=-2\)
So \(x=4, x=-2\)
