n^2-2n-4=0 how do I get this into radical form?

I think you are needing the quadratic formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a = 1    b = -2    c  = -4       Substitute these values to find the values of 'n' that solve the equation    ( I know ....it says 'x'...but you can put 'n' there)