Given:# color(white)(....)y=1/3x^2+5/6x+7/8#

Consider the part inside the brackets:#color(white)(....)y=(1/3x^2+5/6x)+7/8#

Write as: #1/3(x^2 + {5/6 -: 1/3}x)#

#1/3(color(red)(x^2)+color(blue)(5/2color(green)(x)))#

If we halve #5/2# we get #5/4#

Change the bracketed bit so that have

#1/3(color(red)(x)+color(blue)(5/4))^2#

We have changed #color(red)(x^2)# to just #color(red)(x)#; halved the coefficient of #color(green)(x)-> color(blue)(1/2 xx 5/2=5/4)# and totally removed the single #color(green)(x)#

So we know write the equation as:

#y-> 1/3(x+5/4)^2+7/8#

The thing is; we have introduced an error that results from squaring the bracket. The error is when we square the #(+5/4)# bit. This error means that the right no longer ='s the left. That is why I have used #y->#

#color(blue)("To correct for this we write:")#

#y-> 1/3(x+5/4)^2color(blue)(-(5/4)^2)+7/8#

The correction now means that the #color(red)("left does = right.")#

#ycolor(red)(=) 1/3(x+5/4)^2color(blue)(-(5/4)^2)+7/8#

So the arithmetic now gives:

#y=1/3(x+5/4)^2-11/16#