Question

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

Answer 1

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