Question 1015551
<pre>
{{{3x^2+4y^2-6x-24y+39=0}}}

This appears to be the equation of an ellipse:

{{{3x^2-6x+4y^2-24y=-39}}}

{{{3(x^2-2x)+4(y^2-6y)=-39}}}

Complete the square in the first parentheses on the left:
Coefficient of x is -2
Multiply it by 1/2, get -1
Square that: (-1)<sup>2</sup> = +1
Add +1-1 inside the first parentheses on the end:

{{{3(x^2-2x+1-1)+4(y^2-6y)=-39}}}

Complete the square in the second parentheses on the left:
Coefficient of y is -6
Multiply it by 1/2, get -3
Square that: (-3)<sup>2</sup> = +9
Add +9-9 inside the second parentheses on the end:

{{{3(x^2-2x+1-1)+4(y^2-6y+9-9)=-39}}}

Factor the first three terms in each parentheses as
the square of a binomial:

{{{3((x-1)^2-1)+4((y-3)^2-9)=-39}}}

Distribute the 3 and 4 into the outer parentheses 
leaving the two squares of binomials intact:

{{{3(x-1)^2-3+4(y-3)^2-36=-39}}}

{{{3(x-1)^2+4(y-3)^2-39=-39}}}

{{{3(x-1)^2+4(y-3)^2=0}}}

Normally we don't get a 0 on the right, and we divide
through by what we get.  But this is a rare case.  We
cannot divide by 0.  

Both terms are non-negative, so the left side must be 0.  
It can only be 0 if x=1 and y=3.

So this is the equation of the point (1,3).  It is really
an ellipse so small it has shrunk down to a point!

Here is the graph (Yes, that's all there is to it! One point!)

{{{drawing(300,300,-2,5,-2,5,
graph(300,300,-2,5,-2,5),locate(1,3,"(1,3)"),
circle(1,3,0.03),circle(1,3,0.01) )}}}

Edwin</pre>