Rezolvați b=frac{b^2+sqrt{4ac}}{2a}

$$b = \frac { b ^ { 2 } + \sqrt { 4 a c } } { 2 a }$$

Rezolvați pentru a (complex solution)

$$\left\{\begin{matrix}a=-\frac{-b^{3}+\sqrt{c\left(2b^{3}+c\right)}-c}{2b^{2}}\text{; }a=\frac{b^{3}+\sqrt{c\left(2b^{3}+c\right)}+c}{2b^{2}}\text{, }&b\neq 0\text{ and }\left(c=0\text{ or }arg(ab-\frac{b^{2}}{2})<\pi \right)\\a\neq 0\text{, }&b=0\text{ and }c=0\end{matrix}\right,$$

Rezolvați pentru b (complex solution)

$$b=-\sqrt{a^{2}-2\sqrt{ac}}+a$$

$$b=\sqrt{a^{2}-2\sqrt{ac}}+a\text{, }a\neq 0$$

Rezolvați pentru a

$$\left\{\begin{matrix}a\neq 0\text{, }&b=0\text{ and }c=0\\a=-\frac{-b^{3}+\sqrt{c\left(2b^{3}+c\right)}-c}{2b^{2}}\text{, }&\left(c=0\text{ and }b\neq 0\right)\text{ or }\left(c\leq 0\text{ and }b\leq -\sqrt[3]{c}\text{ and }b<0\text{ and }b<-\frac{2^{\frac{2}{3}}}{2}\sqrt[3]{c}\right)\\a=\frac{b^{3}+\sqrt{c\left(2b^{3}+c\right)}+c}{2b^{2}}\text{, }&\left(c\geq 0\text{ and }b\geq -\sqrt[3]{c}\text{ and }b>0\text{ and }b>-\frac{2^{\frac{2}{3}}}{2}\sqrt[3]{c}\right)\text{ or }\left(b\neq 0\text{ and }c=0\right)\end{matrix}\right,$$

Rezolvați pentru b

$$b=-\sqrt{a^{2}-2\sqrt{ac}}+a$$

$$b=\sqrt{a^{2}-2\sqrt{ac}}+a\text{, }\left(a\geq \sqrt[3]{4c}\text{ and }c\geq 0\text{ and }a>0\right)\text{ or }\left(a\leq \sqrt[3]{4c}\text{ and }c\leq 0\text{ and }a<0\right)\text{ or }\left(a=\sqrt[3]{4c}\text{ and }c\neq 0\right)$$