(x-1)(x-3)(x-5)(x-7)=-15
[(x-1)(x-7)][(x-3)(x-5)]=-15
[tex] (x^{2} -8x+7)( x^{2} -8x+15)=-15 \\ notam \\ x^{2} -8x=t \\ (t+7)(t+15)=-15 \\ t^{2}+22t+120=0 [/tex]
Δ=4
[tex] t_{1}= \frac{-22-2}{2}=-12; t_{2}= \frac{-22+2}{2}=-10 [/tex]
revenim la substitutie
[tex] x^{2} -8x=-12; x^{2} -8x=-10\\ x^{2} -8x+12=0; x^{2} -8x+10=0 \\ x=2;x=6;x=[tex]4- \sqrt{6} [/tex];x=[[tex]4+ \sqrt{6} [/tex]/tex]
S={2;6;4-√6;4+√6}
