When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
Sejam \((X_0,\mathcal{O}_{X_0})\) e \((X_1,\mathcal{O}_{X_1})\) dois LRS e \(U_0\subset X_0\), \(U_1\subset X_1\) abertos tais que há um isomorfismo
$$
(\phi,\phi^{\sharp}):(U_0,\mathcal{O}_{X_0}|_{U_0}) \tilde{\longrightarrow} (U_1,\mathcal{O}_{X_1}|_{U_1})
$$
