Answer
{$1\pm\sqrt 3$}
Work Step by Step
Step 1: Comparing $x^{2}-2x-2=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=1$, $b=-2$ and $c=-2$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt (b^{2}-4ac)}{2a}$
Step 3: Substituting the values of a,b and c in the formula:
$x=\frac{-(-2) \pm \sqrt ((-2)^{2}-4(1)(-2))}{2(1)}$
Step 4: $x=\frac{2 \pm \sqrt (4+8)}{2}$
Step 5: $x=\frac{2 \pm \sqrt (12)}{2}$
Step 6: $x=\frac{2 \pm \sqrt (4\times3)}{2}$
Step 7: $x=\frac{2 \pm2\sqrt 3}{2}$
Step 8: $x=\frac{2(1 \pm\sqrt 3)}{2}$
Step 9: $x=1\pm\sqrt 3$
Step 10: $x=1+\sqrt 3$ or $x=1-\sqrt 3$
Step 11: Therefore, the solution set is {$1\pm\sqrt 3$}.
