"Solve y^2 - |4y - 2| + 5 = 0." I'm going to assume only real solutions are required.
(1) Suppose 4y-2 is positive. Then we have y2-4y+7=0 but b2-4ac = 16 - 28 <0 so solutions are complex.
(2) Suppose 4y-2 is negative. Then we have y2+4y+3=0. so y = (-4±√(16-12))/2 or y = -1 and y = -3
If complex solutions are required as well, just solve the quadratic in (1).
I agree with your real solutions, Alan
I also thought that the complex solutions to y^2 - 4y + 7 = 0 were answers
But...Wolframalpha gives these answers for the complex solutions :
Do you see how they got those ???
Hmm! My simplification when |4y-2| is positive leads to incorrect solutions. i.e. when the solutions are plugged back into the LHS of the original equation the resut isn't zero. The Wolfram Alpha solutions, on the other hand, do give zero, so they are obviously correct. I will need to think further about how to obtain these!
Further thought has produced the following:
I guess WolframAlpha used something like this reasoning!