What are the solutions?

Question

haleyelizabeth2017 · Answer

$1/2x^2+2x+3=0$ @Mirahalfian

haleyelizabeth2017 · Answer

Is it $x=-2\pm\sqrt6$ ?

haleyelizabeth2017 · Answer

@TheSmartOne sorry to bug you, but would you mind checking this one?

anonymous · Answer

If in the form of Vertex,

haleyelizabeth2017 · Answer

I don't get how you got that...I found a mistake in mine but fixed it now: 
$x=-2\pm\sqrt-2$ but I don't know how to simplify it any more.....I don't need it in vertex form though........

anonymous · Answer

Hmmm if you got that answer, then the only way to simplify it is by using the "complex number" form that involves in "i" form.

haleyelizabeth2017 · Answer

okay....so would it be $x=-2\pm i\sqrt2$?

haleyelizabeth2017 · Answer

I have no idea...or $x=-2\pm 2i$?

anonymous · Answer

x =  - 2 ± (√ 2) i

haleyelizabeth2017 · Answer

what does that say? I can't see it....all I see is diamonds

anonymous · Answer

Sorry, 
x = -2 +- (square root of 2) i

haleyelizabeth2017 · Answer

Okay I see it now...but why would it be $(\sqrt 2)i$?

haleyelizabeth2017 · Answer

why wouldn't it just be 2i since it is $\sqrt-2$?

anonymous · Answer

i is equal to square root of -1

Yours is Square root of -2,
So, (Square root of 2)*(Square root of -1)
That makes it to be Square root of 2)i

haleyelizabeth2017 · Answer

I'm really confused :/

haleyelizabeth2017 · Answer

but wouldn't the $\sqrt-1$ aka i*2 just equal $\sqrt-2$ I am just utterly confused...