To find the vertex form of the equation, we have to complete the square:
#y=2x^2-5x-3#
#y=(2x^2-5x)-3#
#y=2(x^2-5/2x)-3#
In #y=ax^2+bx+c#, #c# must make the bracketed polynomial a trinomial. So #c# is #(b/2)^2#.
#y=2(x^2-5/2x+((5/2)/2)^2-((5/2)/2)^2)-3#
#y=2(x^2-5/2x+(5/4)^2-(5/4)^2)-3#
#y=2(x^2-5/2x+25/16-25/16)-3#
Multiply #-25/16# by the vertical stretch factor of #2# to bring #-25/16# outside of the brackets.
#y=2(x-5/4)^2-3-((25/16)*2)#
#y=2(x-5/4)^2-3-
((25/color(red)cancelcolor(black)16^8)*color(red)cancelcolor(black)2)#
#y=2(x-5/4)^2-3-25/8#
#y=2(x-5/4)^2-49/8#
