Question

(-7f-2)^2-(5f+4)^2-4(6f^2-5)=-28
решите пример!

Answer 1

$(-7f-2)^2-(5f+4)^2-4(6f^2-5)=-28 \ 49f^2+28f+4-25f^2-40f-16-24f^2+20+28=0 \ -12f+36=0 \ -12f=-36 \ 12f=36 \ f=3$