Question

решите неравенство :
. 2
(y-5)(y+2)-(y-7) >18​

Answer 1

$(y-5)(y+2)-(y-7)>18\\\\y^2-3y-10-y+7>18\\\\y^2-4y-21>0\\\\\star \; \; y^2-4y-21=0\; \; \to \; \; \; y_1=-3\; ,\; y_2=7\; \; (teorema\; Vieta)\; \; \; \Rightarrow \\\\(y+3)(y-7)>0\\\\znaki:\; \; \; +++(-3)---(7)+++\\\\x\in (-\infty ,-3)\cup (7,+\infty )$