## Elementary Algebra

The numbers are: $5 + \sqrt{3}$ and $5 - \sqrt3$
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 $5 - \sqrt3$