Answer
The numbers are: $ 5 + \sqrt{3} $ and $ 5 - \sqrt3$
Work Step by Step
We call one number x and the other y. Thus, we obtain:
$x+y = 10 \\ xy = 22$
Plugging in $x = 10-y$ into equation two, we obtain:
$ (10-y)y=22 \\ y^2 -10y + 22 = 0 $
We apply the quadratic formula:
$ x = \frac{ -b \pm \sqrt{b^2 -4ac}}{2a}$
$ x = \frac{10 \pm \sqrt{(-10)^2 -4(1)(22)}}{2}$
$ x = \frac{ 10 \pm \sqrt{12}}{2}$
$ x = 5 \pm \sqrt3$
The numbers are $ 5 + \sqrt{3} $ and $ 5 - \sqrt3$
