Alternativer Weg:
\( 2 x \cdot e^{-4 x}+\left(x^{2}-3\right) \cdot e^{-4 x} \cdot(-4)=0 \)
\( \frac{ 2 x }{e^{4 x}}+\frac{12-4x^2}{e^{4 x}}=0 |\cdot e^{4 x}\)
\( 2x-4x^2=-12\)
\( x^2-\frac{2}{4}x=3\) quadratische Ergänzung :
\( x^2-\frac{2}{4}x+(\frac{1}{4})^2=3+(\frac{1}{4})^2\) 2.Binom:
\( (x-\frac{1}{4})^2=\frac{49}{16}|±\sqrt{~~}\)
1.)
\( x-\frac{1}{4}=\frac{7}{4}\)
\( x_1=2\)
2.)
\( x-\frac{1}{4}=-\frac{7}{4}\)
\( x_2=-1,5\)