CPhill

Maybe he's "auditioning".........

Sorry, mchow.....there may be a "trick," but I don't know what it might be.....I certainly don't see it....!!!

If something "hits me" later on, I'll message you.......(maybe some other answerer might see something I don't....)

3x^2 - 7x + 12 = 0    this doesn't factor......using the onsite solver, we have

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{95}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{95}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{7}}}{{\mathtt{6}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.624\: \!465\: \!724\: \!134}}{i}\right)\\
{\mathtt{x}} = {\frac{{\mathtt{7}}}{{\mathtt{6}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.624\: \!465\: \!724\: \!134}}{i}\\
\end{array} \right\}$$

The sum of the roots =  (14 / 6) =( 7 / 3 )

The product of the roots is

(1/36)[(7 - (√95)i] [ (7 + (√95)i ] =

(1/36) [49 - 95i²] =

(1/36) [ 49 + 95] =

(1/36)[144] =  4

NO!!!....a negative exponent just indicates a reciprocal....for instance......

3² = 9    but..... 3^-2 = 1 / 3² = 1 / 9

Note that an answer could be negative but not because of the negative exponent itself...for instance

(-3)³  = -27   and......  (-3)^-3  = 1 / (-3)³  = 1 / -27 = -1 / 27

-5 and -7 seem to work OK........

Add the numbers and divide by 10.......

You can use the onsite calculator to evaluate this....

Only when you laugh........

YEAH....BRITTANYJ....!!!!....Finally....a team we can agree on !!!!!!....Steeler fan for life....!!