Question 1067958
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<pre>
Introduce new variable x = (2 -y)^2

The equation becomes

x^2 - 3x - 1 = 0

whose solutions are 

{{{x[1,2]}}} = {{{(3 +- sqrt(13))/2}}}.


Hence, {{{2 - y}}} = +/- {{{sqrt((3 +- sqrt(13))/2)}}},


{{{y}}} = {{{2 +- sqrt((3 +- sqrt(13))/2)}}}.
</pre>

Two of four roots with the sign "-" (minus) under the square radical are complex numbers.