Respuesta :

naǫ
[tex]x^2+2x+9=0 \\ \\ a=1 \\ b=2 \\ c=9 \\ b^2-4ac=2^2-4 \times 1 \times 9=4-36=-32 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-2 \pm \sqrt{-32}}{2 \times 1}=\frac{-2 \pm \sqrt{-16 \times 2}}{2}=\frac{-2 \pm 4i\sqrt{2}}{2}=-1 \pm 2i\sqrt{2}[/tex]

The answer is D.
Benjamin16
[tex]x^{2}+2x+9=0 \\ \\ x^{2}+2x+1+8=0 \\ \\ (x+1)^{2}+(2 \sqrt{2})^{2}=0 \\ \\ (x+1+2 \sqrt{2}i)(x+1-2 \sqrt{2}i)=0 \\ \\ x+1+2 \sqrt{2}i=0 \ \vee \ x+1-2 \sqrt{2}i=0 \\ \\ x=-1-2 \sqrt{2}i \ \vee \ x=-1+2 \sqrt{2}i [/tex]

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[tex]a^{2}+b^{2}=(a+bi)(a-bi)[/tex]

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