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)
+37170 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)
+37170 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)
