Question

Решите неравенства (12+x)^2 ≥ x^2+21​

Answer 1

$(12+x)^2geq x^2+21\144+24x+x^2-x^2geq 21\24xgeq 21-144\24xgeq -123 /:24\xgeq -frac{41}8}\xin[-frac{41}{8};+infty)$