Solve x^2 + 2x + 9 = 0

Help!

Hello, this one is kind of painul because it involves a complex number solution...

First, it's simplest to just complete the square...(Or use the quadratic method above)

We note that

(x+1)(x+1) = X^2 + 2x + 1
So...
(x+1)^2 + 8 = x^2 + 2x + 9 = 0
So...
(x+1)^2 = - 8
x+1 = +/- sqrt(-8)
x = +/- sqrt(-8) - 1

To confirm this is the painful part...I will only confirm one of the two solutions...

[sqrt(-8)-1]^2 + 2*[sqrt(-8)+1] + 9 = 0 [The ugly part as you see...]
[-8 -2*sqrt(-8) + 1] +[2*sqrt(-8)-2] + 9 = 0 [To square a complex number you do the same steps as squaring a binomal]
-8 + 1 -2 + 9 = 0 [We cancel out 2*sqrt(-8)]
0 = 0 [Yep, that works]

a=1 b=2 c=9


quadratic formula: [-b(+/-)sqrt(b^2-4ac)]/2a


[-2(+/-)sqrt(4-36)]/2



[-2(+/-)sqrt(-32)]/2

[-2(+/-)sqrt(32)i]/2

[-2(+/-)2sqrt8i]/2

answer: -1(+/-)sqrt8i.

i is an imaginary number.

x^2 + 2x = -9

x^2 = -2x - 9

x = SQ.Root (-2x - 9)

There you go :D

(Sq. Root is equal to the Square Root symbol btw. I just cant place it on the comp