Answer:
The roots of the given equation are [tex]1+\sqrt{3}[/tex] and [tex]1-\sqrt{3}[/tex].
Step-by-step explanation:
If a quadratic equation is defined as
[tex]ax^2+bx+c=0[/tex]
then according to the quadratic formula the roots of the function are
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The given equation is
[tex]x^2-2x-2=0[/tex]
Here, a=1,b=-2 and c=-2.
Using quadratic formula we get,
[tex]x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-2)}}{2(1)}[/tex]
[tex]x=\frac{2\pm \sqrt{4+8}}{2}[/tex]
[tex]x=\frac{2\pm \sqrt{12}}{2}[/tex]
[tex]x=\frac{2\pm 2\sqrt{3}}{2}[/tex]
[tex]x=1\pm \sqrt{3}[/tex]
Therefore the roots of the given equation are [tex]1+\sqrt{3}[/tex] and [tex]1-\sqrt{3}[/tex].
