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

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Answer 1 · 2015-09-11T15:53:25

x= -1+i; -1-i

Simplify the equation as #x^2 +2x+2=0#

This can be solved by completing the square #x^2+2x+1= -1#

#(x+1)^2 = -1#

x+1= #+- i#

x= -1+i; -1-i

Answer 2 · 2015-09-11T15:55:43

There are no solutions in the set of real numbers

We have that

#3x^2+6x+6=0=>3(x^2+2x+1)+3=0=>3(x+1)^2+3=0#

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