<PRE><B>Question 871053</B>
Simplify first.
{{{3x^2+5x-5=0}}}.  The function would be something general as {{{y=3x^2+5x-5=0}}}.  The term to complete the square is (5/2)^2, which you should ..... Not that.... but better, this...


{{{y=3(x^2+(5/3)x+5/3)=0}}}.  Now, using this might be better.  The term to complete the square is {{{(5/(2*3))^2=(5/6)^2}}}.


{{{y=3(x^2+(5/3)x+(5/6)^2+5/3-(5/6)^2)}}}
{{{y=3((x+5/6)^2+5/3-(5/6)^2)}}}
{{{y=3(x+5/6)^2+3(5/3-(5/6)^2)}}}
{{{y=3(x+5/6)^2+3(60/36-25/36)}}}
{{{y=3(x+5/6)^2+3(35/36)}}}
{{{highlight(y=3(x+5/6)^2+35/12)}}}.  STANDARD FORM.


The vertex is a minimum, at (-5/6, 35/12).  The roots will be found simply solving y=0.  This is partly done already because the equation is now in standard form.</PRE>