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

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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)

Apr 25, 2019
edited by Guest  Apr 25, 2019
edited by Guest  Apr 25, 2019

#1
+19913
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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)

Apr 25, 2019

#1
+19913
+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