How do you solve #m^2 + m + 1 = 0 # using the quadratic formula?

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Answer 1 · 2018-07-09T02:57:23

#m=(-1+sqrt(3)i)/2#, #(-1-sqrt(3)i)/2#

Solve:

#m^2+m+1=0#

This is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=1#, #b=1#, #c=1#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute #m# for #x#. Plug in the known values.

#m=(-1+-sqrt(1^2-4*1*1))/(2*1)#

#m=(-1+-sqrt(-3))/2#

Simplify.

#m=(-1+-sqrt(3)i)/2#

#m=(-1+sqrt(3)i)/2#, #(-1-sqrt(3)i)/2#