Question

Решите уравнение
(х-2)^2+24=(2+3х)^2

Answer 1

$(x-2)^2+24=(2+3x)^2\x^2-4x+4+24=4+12x+9x^2\x^2-4x+4+24-4-12x-9x^2=0\-8x^2-16x+24=0|:(-8)\x^2+2x-3=0\D=2^2-4*1*(-3)=4+12=16\\x_{1}= frac{-2+4}{2} = frac{2}{2} =1\\x_{2}= frac{-2-4}{2} = frac{-6}{2} =-3\\Otvet:1,-3$