An algebra problem by Lloyd Regalado

Let the two numbers be $x$ and $y$ .
According to Question.
$\Rightarrow x+y=12$ ..... $(1)$
$\Rightarrow x×y=32$ ..... $(2)$
From $(1)$ .
$y=12-x$
Putting $y$ in $(2)$ .
$\Rightarrow x×(12-x)=32$
$=12x-x^2=32$
$=x^2-12x+32=0$
$x=4,8$
Putting in $(1)$ .
This gives $x=4$ and $y=8$
Now, larger number is $\boxed{8}.$