+4 Is there a way to define a variable to use an equation? E.x.: 3x^2-2x+5=y, and I want to assign x a value without changing it out. Is it possible?
+118608 Welcome to web2.0calc forum Catboy13
Umm - Interesting question. 
$$\begin{array}{rll}
3x^2-2x+5&=&y\\\\
3x^2-2x&=&y-5\\\\
x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\
x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\
x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\
\end{array}$$
If you first define a function f as
$${f}{\left({\mathtt{x}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}$$
then for any particular x the function f(x) has a value and you can write things such as
f(2) + f(0) - 4 = 14
+118608 Welcome to web2.0calc forum Catboy13
Umm - Interesting question.
$$\begin{array}{rll}
3x^2-2x+5&=&y\\\\
3x^2-2x&=&y-5\\\\
x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\
x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\
x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\
\end{array}$$
