Let's call the two numbers x and y. The first sentence, one is 6 less than twice another, translates to
[tex] x = 2y-6 [/tex]
We also know the sum of their squares:
[tex] x^2+y^2 = 680 [/tex]
So, we have the following system:
[tex] \begin{cases} x = 2y-6 \\ x^2+y^2 = 680 \end{cases} [/tex]
We can use the expression for x in terms of y from the first equation to turn the second equation into something involving y alone:
[tex] x^2+y^2 = 680 \to (2y-6)^2+y^2 = 680 [/tex]
Expand the square:
[tex] 5y^2-24y+36 = 680 [/tex]
Subtract 680 from both sides:
[tex] 5y^2-24y-644 = 0 [/tex]
Using the quadratic equation
[tex] y_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
we find the two solutions
[tex] y_1 = -\cfrac{46}{5},\quad y_2 = 14 [/tex]
Since we know that both numbers are integer we can only accept the second solution. It yields the following x value:
[tex] x = 2x-6\ \land\ y = 14 \implies x = 2\cdot 14 - 6 = 28-6=22 [/tex]
