How do you solve the quadratic #3y^2+3y+1=0# using any method?

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Answer 1 · 2016-12-17T19:19:25

The answer is # S={-1/2-isqrt3/6,-1/2+isqrt3/6}#

If you have a quadratic equation

#ax^2+bx+c=0#

#x=(-b+-sqrtDelta)/(2a)#

Here our equation is

#3y^2+3y+1=0#

You calculate the discriminant

#Delta=b^2-4ac#

#=9-4*3*1=-3#

As, #Delta<0#, there are no solutions in #RR# but in #CC#

The solutions are

#y=(-3+-sqrtDelta)/(2*3)=(-3+-isqrt3)/6#

#y_1=-1/2-isqrt3/6#

and #y_2=-1/2+isqrt3/6#