Question 48519
Two ways of doing this, I think the second one will do better.
First way:
a trigonomial can be written like ax^2+bx+c
f(x)=-1x^2+6xy+9y^2
a=-1
b=6y
c=9y^2
if f(x)=0
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
{{{x=3y(1+-sqrt(2))}}}
{{{f(x)=-(x-3y(1+sqrt(2)))(x-3y(1-sqrt(2)))}}}
Second way:
{{{f(x)=-1x^2+6xy+9y^2}}}
={{{18y^2-(x^2-6xy+9y^2)}}}
={{{((3sqrt(2))y-(x-3y))((3sqrt(2))y+(x-3y))}}}
={{{((3sqrt(2))y-(x-3y))((3sqrt(2))y+(x-3y))}}}
={{{(3y(sqrt(2)+1)-x)(3y(sqrt(2)-1)+x)}}}