Question

Помогите решить уравнение

Answer 1

$4^{5x^{2}+3x }-20=-32^{x^{2}+\frac{3x}{5}}\\\\4^{5x^{2}+3x }-20=-(2^5)^{x^{2}+\frac{3x}{5}}\\\\4^{5x^{2}+3x }-20=-2^{5x^{2}+3x}\\\\(2^{2})^{5x^{2}+3x }+2^{5x^{2}+3x}-20=0\\\\2^{5x^{2}+3x}=m, \ m>0\\\\m^{2}+m-20=0\\\\m_{1}=4\\\\m_{2}=-5<0-neyd\\\\2^{5x^{2}+3x}=4\\\\2^{5x^{2}+3x}=2^{2}\\\\5x^{2}+3x=2\\\\5x^{2}+3x-2=0\\\\D=3^{2}-4*5*(-2)=9+40=49=7^{2}\\\\x_{1} =\frac{-3-7}{10}=-1\\\\x_{2}=\frac{-3+7}{10}=0,4\\\\Otvet:\boxed{-1 \ ; \ 0,4}$