(x+3)(x-5)(x+2)(x-4)=60

(x+3)(x-5)(x+2)(x-4)=60
$( x^{2} -5x+3x-15)*( x^{2} -4x+2x-8)=60 \\ ( x^{2} -2x-15)*( x^{2} -2x-8)=60$
Пусть $x^{2} -2x-15= \alpha$ . Тогда
$\alpha ( \alpha +7)=60 \\ \alpha^{2} +7 \alpha -60=0 \\ \alpha_{1} =-12; \alpha _{2} =5 \\ \\ x^{2} -2x-15=-12 \\ x^{2} -2x-3=0 \\ x_{1} =-1; x_{2}=3 \\ \\ x^{2} -2x-15=5 \\ x^{2} -2x-20=0 \\ x_{1}=1- \sqrt{21}; x_{2}=1+ \sqrt{21}$