20-2q^2=40+2q

20-2q^2=40+2q       Add 2q^2 to both sides and subtract 20 from both sides

0 = 2q^2 + 2q + 40 - 20    Rewrite as

2q^2 + 2q + 20  = 0     Divide everything by 2    

q^2 + q +10 = 0    This won't factor, so using the onsite calculator to find the solutions, we have

$${{\mathtt{q}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{q}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{q}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{q}} = {\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{q}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.122\: \!498\: \!999\: \!200\: \!543\: \!4}}{i}\right)\\
{\mathtt{q}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.122\: \!498\: \!999\: \!200\: \!543\: \!4}}{i}\\
\end{array} \right\}$$

So this has no "real" solutions......