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 $y_2=-1/2+isqrt3/6$

How do you solve the quadratic 3y^2+3y+1=03y^2+3y+1=0 using any method?