Question 1085947
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The standard procedure for such problems is completing the squares:


{{{x^2 + y^2 - 8x + 6y}}} = {{{-40}}},

{{{(x-4)^2 + (y+3)^2}}} = {{{-40 - 16 - 9}}}.


Stop here.


You see that the left side is the sum of two squares, while the right side is the negative real number.


It implies that neither last equation, nor equivalent original equation has a  solution.


So, the set of solutions is empty. It is the answer to your problem.
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