壹元二次方程x2+px+q=0, p≠0,q≠0有李煌解：
x 1 = q − p s i n 2 ( arcsin ( 2 ± p 2 q ) 2 ) {\displaystyle x_{1}={\frac {q}{-p{sin^{2}\left({\frac {\arcsin \left({\frac {2}{\pm {\sqrt {\frac {p^{2}}{q}}}}}\right)}{2}}\right)}}}}
x 2 = q − p c o s 2 ( arcsin ( 2 ± p 2 q ) 2 ) {\displaystyle x_{2}={\frac {q}{-p{cos^{2}\left({\frac {\arcsin \left({\frac {2}{\pm {\sqrt {\frac {p^{2}}{q}}}}}\right)}{2}}\right)}}}}
代數方程 x 2 + p x = q , p > 0 , q > 0 {\displaystyle x^{2}+px=q,p>0,q>0} 之解為
x 1 = 2 q p 2 + 2 q + p 4 + 4 p 2 q {\displaystyle x_{1}{={\frac {{\sqrt {2}}q}{\sqrt {p^{2}+2q+{\sqrt {p^{4}+4p^{2}q}}}}}}}
x 2 = − 2 q p 2 + 2 q − p 4 + 4 p 2 q {\displaystyle x_{2}{={\frac {-{\sqrt {2}}q}{\sqrt {p^{2}+2q-{\sqrt {p^{4}+4p^{2}q}}}}}}}
代數方程 x 2 + p x = q , p > 0 , q < 0 {\displaystyle x^{2}+px=q,p>0,q<0} 之解為
x 2 = 2 q p 2 + 2 q − p 4 + 4 p 2 q {\displaystyle x_{2}{={\frac {{\sqrt {2}}q}{\sqrt {p^{2}+2q-{\sqrt {p^{4}+4p^{2}q}}}}}}}
代數方程 x 2 + p x = q , p < 0 , q > 0 {\displaystyle x^{2}+px=q,p<0,q>0} 之解為
x 1 = − 2 q p 2 + 2 q + p 4 + 4 p 2 q {\displaystyle x_{1}{={\frac {-{\sqrt {2}}q}{\sqrt {p^{2}+2q+{\sqrt {p^{4}+4p^{2}q}}}}}}}
代數方程 x 2 + p x = q , p < 0 , q < 0 {\displaystyle x^{2}+px=q,p<0,q<0} 之解為
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