Question 937988

I'm trying to figure out how to find "Complex Zeros of Quadratic Functions" but the Holt McDougal video lesson isn't helping whatsoever and I've researched as much as I could. I'm trying to find the zeros of "f(x)=x²+2x+3" and the video says the answer is x=6+2i√3. This involves imaginary numbers.
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{{{f(x) = x^2 + 2x + 3}}}
Using the QUADRATIC EQUATION formula: {{{x = (- b +- sqrt(b^2 - 4ac))/(2a)}}}, and with:
a being 1; b being 2; and c being 3, {{{x = (- b +- sqrt(b^2 - 4ac))/(2a)}}} becomes: 
{{{x = (- 2 +- sqrt((2)^2 - 4(1)(3)))/(2(1))}}}
{{{x = (- 2 +- sqrt(4 - 12))/2}}}
{{{x = (- 2 +- sqrt(- 8))/2}}}
{{{x = (- 2 +- sqrt(- 1 * 2^2 * 2))/2}}}
{{{x = (- 2 +- 2i * sqrt(2))/2}}} ------ {{{sqrt(- 1) = i}}} and {{{sqrt(2^2) = 2}}}
{{{x = (2(- 1 +- i*sqrt(2)))/2}}}
{{{x = (cross(2)(- 1 +- i*sqrt(2)))/cross(2)}}}
{{{highlight_green(x = - 1 +- sqrt(2)*i)}}}
This is the correct answer, but is nowhere close to what you claim the answer is. You need to re-check the problem!!