Задание 701

$$q=93-3p, r(p)=pq \Rightarrow$$

$$r(p)=p(93-3p)=93p-3p^2$$

$$93p-3p^2=660$$

$$93p-3p^2-660=0$$

$$3p^2-93p+660=0$$

$$p^2-31p+220=0$$

$D=(-31)^2-4\cdot 220=$$961-880=81$

$p_1=\frac{-(-31)+\sqrt{81} }{2}=\dfrac{31+9}2=$$\dfrac{40}2=20$

$p_2=\frac{31-\sqrt{81} }{2}=\dfrac{31-9}2=$$\dfrac{22}2=11$

$$20>11$$