[tex]x;\ y-the\ numbers\\\\ \left\{\begin{array}{ccc}x+y=10&\to x=10-y\\x\cdot y=12\end{array}\right\\\\\\subtitute\ x=10-y\ to\ the\ equation\ x\cdot y=12\\\\(10-y)\cdot y=12\\\\10y-y^2=12\ \ \ |subtract\ 12\ from\ bot\ sides\\\\-y^2+10y-12=0\ \ \ |change\ the\ signs\\\\y^2-10y+12=0[/tex]
'[tex]Use\ the\ quadratic\ formula:ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta \ \textgreater \ 0\ therefore\ solutions\ is:\ y_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ y_2=\dfrac{-b+\sqrt\Delta}{2a}[/tex]
[tex]y^2-10y+12=0\\a=1;\ b=-10;\ c=12\\\\\Delta=(-10)^2-4(1)(12)=100-48=52\\\sqrt{\Delta}=\sqrt{52}=\sqrt{4\cdot13}=\sqrt{4}\cdot\sqrt{13}=2\sqrt{13}[/tex]
[tex]y_1=\dfrac{-(-10)-2\sqrt{13}}{2\cdot 1}=\dfrac{10-2\sqrt{13}}{2}=\huge\boxed{5-\sqrt{13}}\\\\y_2=\dfrac{-(-10)+2\sqrt{13}}{2\cdot1}=\dfrac{10+2\sqrt{13}}{2}=\huge\boxed{5+\sqrt{13}}[/tex]
[tex]subtitute\ the\ values\ of\ "y"\ to\ the\ equation\ x=10-y\\\\x_1=10-(5-\sqrt{13})=10-5+\sqrt{13}=\huge\boxed{5+\sqrt{13}}\\\\x_2=10-(5+\sqrt{13})=10-5-\sqrt{13}=\huge\boxed{5-\sqrt{13}}[/tex]
[tex]Your\ answer:\\\boxed{x=5+\sqrt{13}\ and\ y=5-\sqrt{13}\ or\ x=5-\sqrt{13}\ and\ y=5+\sqrt{13}}[/tex]
