0.10 = 6.0/v^2 + 0.00050v^2 - 0.033

$$v\ne0$$

$$0.1=\frac{6}{v^2}+0.0005v^2-0.033\\\\ 0.133=\frac{6}{v^2}+0.0005v^2\\\\ 0.133v^2=6+0.0005v^4\\\\$$

substitute x for v²

$$0.0005x^2-0.133x+6=0\\$$

$${\mathtt{0.000\: \!5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.133}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{133}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5\,689}}}}\\ {\mathtt{x}} = {\sqrt{{\mathtt{5\,689}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{133}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{57.574\: \!540\: \!107\: \!467\: \!690\: \!7}}\\ {\mathtt{x}} = {\mathtt{208.425\: \!459\: \!892\: \!532\: \!309\: \!3}}\\ \end{array} \right\}$$

  $$v^2=57.574540\:\:\rightarrow v=\pm7.58779\:\: \mbox{approx}$$    or

$$v^2=208.42545989\:\:\rightarrow\:\:v=\pm14.4369\:\:\mbox{approx}$$

I think that is all okay.