Question 723678
Here is a way to start.


{{{y=3x^2+2x-8}}}
{{{y=3(x^2+(2/3)x  )-8}}}


The expression, {{{x^2+(2/3)x=x(x+2/3)}}}, represents a rectangle with one dimension x and the other dimension {{{x+2/3}}}.  It may be not a square, but you can ADD a term to it so that it becomes a square.  The term to add is {{{(2/(3*2))^2=1/9}}}.  You also must subtract it in order to maintain equality for y.  


Continuing toward this,
{{{y=3(x^2+(2/3)x+1/9)-3(1/9)-8}}}
{{{y=3(x+1/3)^2-(3(1/9)+8)}}}
Can you continue this?