Find real points satisfying 3x^2+3y^2-4xy+10x-10y+10=0

Given

This can be rewritten as follows:


Notice that this is a quadratic equation of the form:
,
with the solution given by:

where:


Thus,


Now consider

This gives a real solution if and only if

i.e.


Thus,

when x = -1

Thus, the equation has real solutions only when x = -1 at which the value of y is given by


Therefore, the real point satisfying

is x = -1, y = 1