Any idea how to linearize this equation? $X^2-Y^2=aZ+bZ^2$ The intention is to linearize this equation $X^2-Y^2=aZ+bZ^2$ into something which looks like $Z=mX+nY+c$ so that a graph of $Z$ against $X$ or $Y$ can be plotted.
X,Y,Z are variables while a,b,c are constants

EDIT: Thanks to dylan37 for pointing out that the this is probably not the sort of "linearization" that curiousfurious (the OP) intended.
Maybe you could use the trick that's employed in deriving the Quadratic Formula.
Starting from $ X^{2}-Y^{2}=aZ+bZ^{2}$, first divide both sides by $b$ to get
$\frac{1}{b}X^{2}-\frac{1}{b}Y^{2}=\frac{a}{b}Z+Z^{2}=2\left(\frac{a}{2b}\right)Z+Z^{2}.$
Now, complete the square:
$
\frac{1}{b}X^{2}-\frac{1}{b}Y^{2}+\frac{a^{2}}{4b^{2}}=\frac{a^{2}}{4b^{2}}+2\left(\frac{a}{2b}\right)Z+Z^{2}=(Z+\frac{a}{2b})^{2}
$
Finally, let$\;\;U=X^{2};\;\;V=Y^{2};\;\;W=(Z+\frac{a}{2b})^{2}.$
In this way, 
$X^{2}-Y^{2}=aZ+bZ^{2}$ becomes $\frac{1}{b}U-\frac{1}{b}V+\frac{a^{2}}{4b^{2}}=W$, or $W=\frac{1}{b}U-\frac{1}{b}V+\frac{a^{2}}{4b^{2}}$.
