Question 279767
{{{3y^2+6y-x+6=0}}}
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Standard form is

{{{(y-k)^2=4p(x-h)}}}

Get the y-terms on the left and the others on the right:

{{{3y^2+6y-x+6=0}}}

{{{3y^2+6y=x-6}}}

Divide through by 3, the coefficient of {{{y^2}}}:

{{{y^2+2y=1/3}}}{{{x-2}}}

Complete the square on the left:

1. Multiply the coefficient of y, which is 2 by {{{1/2}}}, getting 1
2. Square 1, getting 1
3. Add 1 to both sides:

{{{y^2+2y+red(1)=1/3}}}{{{x-2+red(1)}}}

Factor the left side as a perfect square, combine terms on the right.

{{{(y+1)^2=1/3}}}{{{x-1}}}

Factor out {{{1/3}}}, the coefficient of x on the right,
by dividing the {{{-1}}} by {{{1/3}}} like this:  
{{{-1}}}{{{"÷"}}}{{{1/3}}}{{{"="}}}{{{-1}}}{{{""*""}}}{{{3/1}}}{{{"="}}}{{{-3}}}

So we have the standard form:

{{{(y+1)^2=1/3}}}{{{(x-3)}}}

Edwin</pre>