Wir setzen

${\displaystyle {}Y=2X^{2}-2\,}$

in die Kreisgleichung ein und erhalten

${\displaystyle {}{\begin{aligned}0&=X^{2}+Y^{2}-1\\&=X^{2}+{\left(2X^{2}-2\right)}^{2}-1\\&=X^{2}+4X^{4}-8X^{2}+4-1\\&=4X^{4}-7X^{2}+3.\end{aligned}}}$

Dies führt auf

${\displaystyle {}X^{4}-{\frac {7}{4}}X^{2}+{\frac {3}{4}}=0\,}$

und somit auf

${\displaystyle {}{\left(X^{2}-{\frac {7}{8}}\right)}^{2}-{\frac {49}{64}}+{\frac {3}{4}}=0\,.}$

Daher ist

${\displaystyle {}{\begin{aligned}X^{2}&=\pm {\sqrt {{\frac {49}{64}}-{\frac {3}{4}}}}+{\frac {7}{8}}\\&=\pm {\sqrt {{\frac {49}{64}}-{\frac {48}{64}}}}+{\frac {7}{8}}\\&=\pm {\frac {1}{8}}+{\frac {7}{8}},\end{aligned}}}$

also

${\displaystyle {}X^{2}=1\,}$

oder

${\displaystyle {}X^{2}={\frac {3}{4}}\,.}$

Somit ist

${\displaystyle {}X=1,-1,{\frac {\sqrt {3}}{2}},-{\frac {\sqrt {3}}{2}}\,.}$