Question 657487
Let's call the two numbers {{{x}}} and {{{y}}}.
The problem tells us that {{{x+y=3}}} and {{{x^2+y^2=29}}}.
 
{{{x+y=3}}} --> {{{(x+y)^2=3^2}}} --> {{{x^2+y^2+2xy=9}}}
Subtracting {{{x^2+y^2=29}}} from {{{x^2+y^2+2xy=9}}}, we get.
{{{x^2+y^2+2xy-x^2-y^2=9-29}}} --> {{{2xy=-20}}} --> {{{xy=-10}}}
 
We can find number that satisfy {{{xy=-10}}} and {{{x+y=3}}}.
We see that {{{highlight(5)}}} and {{{highlight(-2)}}} work.
 
We could also set up and solve a quadratic equation
{{{xy=-10}}} --> {{{y=-10/x}}} (dividing both sides by {{{x}}})

and substituting into {{{x+y=3}}}, we get
{{{x-10/x=3}}} --> {{{x^2-10=3x}}} (multiplying both sides times {{{x}}})
Then {{{x^2-10=3x}}}  --> {{{x^2-3x-10=0}}}  (subtracting {{{3x}}} from both sides).
Of course, we find the same solutions.
We find {{{x=5}}}, which leads to {{{y=-2}}}
and we find {{{x=-2}}}, which leads to {{{y=5}}}.
Either way, the numbers are {{{highlight(5)}}} and {{{highlight(-2)}}}.
It does not matter which number we call {{{x}}}.