Question

Решите уравнение:
1) 3(x+4)(в квадрате)=10x+32​

Answer 1

$3(x+4)^2=10x+32\\\\\\3(x^2+8x+16)-10x-32=0\\\\\\3x^2+24x+48-10x-32=0\\\\\\3x^2+14x+16=0\\\\\\D=196-4\cdot3\cdot16=196-192=4\\\\\\x_1=\dfrac{-14-2}{6}=\dfrac{-16}{6}=-\dfrac{8}{3}\\\\\\x_2=\dfrac{-14+2}{6}=\dfrac{-12}{6}=-2\\\\\\\boxed{Otvet:\ x_1=-\dfrac{8}{3},\ x_2=-2.}$