Respuesta :
Answer:
[tex](x-(1+\sqrt{3}))(x-(1-\sqrt{3}))[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} -2x-2=0[/tex]
so
[tex]a=1\\b=-2\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{12}} {2}[/tex]
[tex]x=\frac{2(+/-)2\sqrt{3}} {2}[/tex]
[tex]x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}[/tex]
[tex]x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}[/tex]
therefore
[tex]x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))[/tex]