Question

решите уравнение
17x+97=29(x+3)^2
​

Answer 1

$17x+97=29(x+3)^2 \ 17x+97=29(x^2+6x+9) \ 17x+97=29x^2+174x+261 \29x^2+174x+261 -17x-97=0 \ 29x^2+157x+164=0 \ D=157^2-4*29*164=5625=75^2 \ x_1= frac{-157-75}{2*29}= -frac{232}{58}=-4 \ \ x_2= frac{-157+75}{2*29}= -frac{41}{29}=-1 frac{12}{29}$