Question 443754
Let's complete the square here:

{{{x^2 -6x = -13}}}
Remember complete the square by adding {{{(b/2)^2}}} to both sides.
{{{x^2 -6x +9 = -13 + 9}}}
This works because now you can factor the left into {{{(x-(b/2))^2}}} always.
{{{(x-3)^2 = -4}}}
{{{x-3 = sqrt(-4)}}}
{{{sqrt(-4)}}} does not exist in real numbers.
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If you are not working with imaginary numbers, you would say no real solutions:
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If you are allowed to work with them, read on.

{{{x-3  = 0+- 2i}}}
{{{x = 3+- 2i}}}