Question 1065310
The equation 2x^2-5x=-12 is rewritten in the form of 2(x-p)^2+q=0.

<pre>
{{{ 2x^2-5x=-12}}}
{{{ 2x^2-5x+12=0}}}

So we have the identity

{{{2(x-p)^2+q=2x^2-5x+12}}}

Substitute x = 0

{{{2p^2+q=12}}}

{{{q=12-2p^2}}}

Substitute x=p

{{{2(p-p)^2+q=2p^2-5p+12}}}

{{{q=2p^2-5p+12}}}

Equate the two expressions for q

{{{12-2p^2}}} &#8801; {{{2p^2-5p+12}}}

{{{-4p^2+5p}}} &#8801; {{{0}}}

{{{p(-4p+5)}}} &#8801; {{{0}}}

{{{p=0}}} or {{{-4p+5=0}}}
             {{{p=5/4}}}

But p cannot be 0, since

{{{2(x-0)^2+q}}} &#8801; {{{2x^2-5x+12}}}

{{{2x^2+q}}} &#8801; {{{2x^2-5x+12}}} cannot be an identity, since 
there is no x term on the left

{{{q=12-2p^2}}}
{{{q=12-2(5/4)^2}}}
{{{q=12-2(25/16)}}}
{{{q=12-25/8}}}
{{{q=71/8}}}

So (p,q) = {{{(matrix(1,3,5/4,",",71/8))}}}

But the first

Edwin</pre>