First equation : y1 = 2x2 - 12x + 15
\Second equation : y2 = 2(x-3)2 + 3
\Apply formula : (a-b)2=a2-2ab+b2
\y2 = 2( x2 - 2(x)(3) - 32 ) + 3
\y2 = 2( x2 - 6x - 9 ) + 3
\y2 = 2x2 - 12x - 18 + 3
\y2 = 2x2 - 12x - 15
\y2 is not equals to y1
\Solution :
\First equation (y1 = 2x2 - 12x + 15) and Second equation (y2 = 2(x-3)2 + 3) algebraically that they are not equivalent