[tex] \displaystyle\\
x^4-5x^2-36=0\\\\
\text{Facem substitutia: }~~y=x^2\\\\
y^2-5y-36=0\\\\
y_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{5\pm \sqrt{25+144}}{2}=\\\\
=\frac{5\pm \sqrt{169}}{2}=\frac{5\pm 13}{2}\\\\
y_1 = \frac{5+ 13}{2}=\frac{18}{2}=9\\\\
y_2 = \frac{5-13}{2}=\frac{-8}{2}=-4\\\\
\text{Ne intoarcem la substitutie.}\\\\
x^2 = 9\\
x = \pm\sqrt{9} \\
x_1 = 3\\
x_2=-3\\\\
x^2=-4\\
x=\pm\sqrt{-4}\\
x_3 = 2i\\
x_4 = -2i [/tex]
