Question 889913
First put into general form.


{{{2x^2-5x-2=0}}}
Factor the 2.
{{{2(x^2-(5/2)x-1)=0}}}
The term which completes the square is {{{(5/(2*2))^2=(5/4)^2}}}, which you add and subtract inside the grouped expression.
{{{2(x^2-(5/2)x+(5/4)^2-(5/4)^2-1)=0}}}
Factor the square trinomial part.
{{{2((x-(5/4))^2-(5/4)^2-(16/16))=0}}}, combined step.
-
{{{2((x-(5/4))^2-(25+16)/16)=0}}}
{{{2(x-5/4)^2-2(41/16)=0}}}
{{{highlight(2(x-5/4)^2-41/8=0)}}}
This is standard form.