So, we can write it like this:
x+y=3
xy=12
from the first we know that
x+y=3
y=3-x
so:
xy=12 can be written as:
x*(x-3)=12
so x^2-3x=12
Actually this equation does not have a solution within the real numbers
but we can calculate:
x^2-3x-12=0
and if we use the quadratic formula, we will have:
(remember the quadratic formula is the following, where for us a=1,b=-3 and c=-12)
\(\frac{-b- \sqrt{b*b-4a*c} }{2a}\)
\(\frac{3- \sqrt{3*3-4*(-12)} }{2} = \frac{3- \sqrt{9+48} }{2} = \frac{3- \sqrt{57} }{2}\)
So this would be the solution, where the other number would be
\(3-\frac{3- \sqrt{57} }{2}= \frac{6}{2}-\frac{3- \sqrt{57} }{2}=\frac{3- \sqrt{57} }{2}\)
The other solution would be the following one;
x=\(\frac{3+\sqrt{57} }{2}\), y=\(1\frac{1+\sqrt{57} }{2}\)