Question 411318
Find the solution set in a +bi form: 6x^2 + 4x +1 =0
..

6x^2 + 4x +1 =0
Use quadratic formula below:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
with a=6,b=4,c=1
..
(-4+-sqrt(4^2-4*6*1))/2*6
=(-4+-sqrt(16-24))/12
=(-4+-sqrt(-8))/12
=(-4+-2*sqrt(-2))/12
=(-4+-2*sqrt(2)i)/12
=(-2+-sqrt(2)i)/6
=-2/6+-(sqrt(2)/6)i)
ans:
-2/6+(sqrt(2)/6)i)
or
-2/6-(sqrt(2)/6)i)