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

I never learned how to.

never mind i found it.

It is 1+/-sqrt(5)

Guest Apr 25, 2019

edited by
Guest
Apr 25, 2019

edited by Guest Apr 25, 2019

edited by Guest Apr 25, 2019

#1**+2 **

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)

ElectricPavlov Apr 25, 2019

#1**+2 **

Best Answer

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)

ElectricPavlov Apr 25, 2019