f(x)= 2x^3-7x+3; f(x)=1

f(x)= 2x^3-7x+3; f(x)=1

We are trying to find the "x's" that make these are equal, so let's set them equal....

2x^3 - 7x + 3  = 1     subtract 1 from both sides

2x^3 - 7x + 2 = 0     this can't be factored.....using the onsite solver, we have

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = -{\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.292\: \!893\: \!218\: \!813\: \!452\: \!5}}\\
{\mathtt{x}} = {\mathtt{1.707\: \!106\: \!781\: \!186\: \!547\: \!5}}\\
{\mathtt{x}} = -{\mathtt{2}}\\
\end{array} \right\}$$

Note, we could have used the Rational Zeros Theorem to find the first root of -2, and then used the Quadratic Formula to find the other two solutions........