Select the approximate values of x that are solutions to f(x) = 0, where f(x) = -7x2 + 6x + 7.

Since this equation is easy to solve using the quadratic formula (see below), what is your criterion for "approximate"?  Is it just a number of decimal places, or are you expected to guess a few values and home in on the results, or what?

 

$${\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{58}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.659\: \!396\: \!157\: \!980\: \!558\: \!3}}\\
{\mathtt{x}} = {\mathtt{1.516\: \!539\: \!015\: \!123\: \!415\: \!5}}\\
\end{array} \right\}$$

 

To two decimal places the values are x = -0.66 and x = 1.52

.