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

Dec 17, 2016

The answer is $S = \left\{- \frac{1}{2} - i \frac{\sqrt{3}}{6} , - \frac{1}{2} + i \frac{\sqrt{3}}{6}\right\}$

#### Explanation:

If you have a quadratic equation

$a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

Here our equation is

$3 {y}^{2} + 3 y + 1 = 0$

You calculate the discriminant

$\Delta = {b}^{2} - 4 a c$

$= 9 - 4 \cdot 3 \cdot 1 = - 3$

As, $\Delta < 0$, there are no solutions in $\mathbb{R}$ but in $\mathbb{C}$

The solutions are

$y = \frac{- 3 \pm \sqrt{\Delta}}{2 \cdot 3} = \frac{- 3 \pm i \sqrt{3}}{6}$

${y}_{1} = - \frac{1}{2} - i \frac{\sqrt{3}}{6}$

and ${y}_{2} = - \frac{1}{2} + i \frac{\sqrt{3}}{6}$