y=A+(Bx+C)^2=A+B^2x^2+2BCx+C^2 has to be equivalent to 4x^2-2x+5.
Equating terms we have:
x^2: B^2=4, B=±2
x: 2BC=-2, BC=-1 so C=-1/B so C=1/2 if B=-2 and C=-1/2 if B=2
Constant: A+C^2=5, so A=5-C^2, and A=5-1/4=19/4.
SOLUTION: A=19/4, (B=-2, C=1/2) or (B=2, C=-1/2)
y=4x^2-2x+5=19/4+(2x-1/2)^2 or y=19/4+(1/2-2x)^2.