Question

(х²-3х)²+3(х²-3х)-28=0 помогите решить

Answer 1

$(x^2-3x)^2+3(x^2-3x)-28=0\x^2-3x=t,;(x^2-3x)^2=t^2\t^2+3t-28=0\D=9+4cdot28=121\t_1=4,;t_2=-7\x^2-3x=4\x^2-3x-4=0\D=9+4cdot4=25\x_1=4,;x_2=-1\x^2-3x=-7\x^2-3x+7=0\D=9-4cdot7=-19;-;KOPHEu;HET\\OTBET:;x_1=4,;x_2=-1$