Respuesta :

Answer:

Option b.

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

[tex]ax^{2} +bx+c=0[/tex] i

s equal to

[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]

in this problem we have

[tex]2x^{2} +2x+3=0[/tex]  

so

[tex]a=2\\b=2\\c=3[/tex]

substitute in the formula

[tex]x=\frac{-2\pm\sqrt{2^{2}-4(2)(3)}} {2(2)}[/tex]

[tex]x=\frac{-2\pm\sqrt{-20}} {4}[/tex]

remember that

[tex]i=\sqrt{-1}[/tex]

so

[tex]x=\frac{-2\pmi\sqrt{20}} {4}[/tex]

[tex]x=\frac{-2\pm2i\sqrt{5}} {4}[/tex]

simplify

[tex]x=\frac{-1\pm i\sqrt{5}i} {2}[/tex]

therefore

In factored form the quadratic equation is equal to

[tex]2x^{2} +2x+3=2(x-(\frac{-1+i\sqrt{5}} {2}))(x-(\frac{-1-i\sqrt{5}} {2}))[/tex]  

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