Knowing to what degree delta should be in limit definition of multivariable functions

Question

I know that by limit definition, $\forall\epsilon>0, \exists \delta>0$ such that

If $0<||P-A||<\delta$ then $|f(P)-L|<\epsilon$

Now I am answering a question such as

Prove, by definition, that $\lim_{(x,y)\to(2,3)} (3x^2+xy-2y^2)=0$.

The problem is I can't get a wrap around to how much will I increase $\delta$ to find it suitable for $\epsilon$. I get stuck at ${(x-2)}^2+{(y-3)}^2<\delta^2$ and I can't make it to become $(3x^2+xy-2y^2)$.

How should I easily know when should I stop increasing the value of $\delta$ to find an $\epsilon>f(P)$?

Answer 1

Hint: Write $3x^2+xy-2y^2=(x+y)(3x-2y)=(x-2+y-3+5)(3(x-2)-2(y-3))$.