Answer
The numbers are $10$ and $15$.
Work Step by Step
equation 1: $x + y =25$
equation 2: $x^{2} + y^{2} = 325$
From equation 1: $x = 25-y$ substituting this to equation 2: $(25-y)^{2} + y^{2} = 325$
$$625 - 50y + y^{2} + y^{2} = 325$$
Simplify: $2y^{2} - 50y + 300 = 0$
Use the quadratic formula to get the roots: $a=2$, $b=-50$, $c=300$ $$y = \frac{-b±\sqrt{b^2-4ac}}{2a}$$ $$y = \frac{-(-50)±\sqrt{(-50)^2-4(2)(300)}}{2(2)}$$ $$y = 15, y=10$$
Substituting to equation 1: $x + y = 25$
if $y =15$, $x +15 = 25$ $$x = 10$$
if $y=10$, $x + 10 = 25$ $$x = 15$$
