3x^2 + 3x - 3 = 0 x = ?

3x^2 + 3x - 3 = 0    divide everything by 3

x^2 + x - 1  = 0

This cannot be be factored, but the solutions lead to two very important numbers........  - Phi and phi...

Here they are

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$

The negative of the larger number is Phi.......and the smaller number is the reciprocal of this.....

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$