# How do you solve 3x^2 + 6x + 6 = 0?

Sep 11, 2015

x= -1+i; -1-i

#### Explanation:

Simplify the equation as ${x}^{2} + 2 x + 2 = 0$

This can be solved by completing the square ${x}^{2} + 2 x + 1 = - 1$

${\left(x + 1\right)}^{2} = - 1$

x+1= $\pm i$

x= -1+i; -1-i

There are no solutions in the set of real numbers

#### Explanation:

We have that

$3 {x}^{2} + 6 x + 6 = 0 \implies 3 \left({x}^{2} + 2 x + 1\right) + 3 = 0 \implies 3 {\left(x + 1\right)}^{2} + 3 = 0$

From the last result we conclude that the equation does not have any real roots.