Задание 3170

$$q=232-4p, r(p)=pq\Rightarrow$$

$$r(p)=p(232-4p)=232p-4p^2$$

$$3300=232p-4p^2$$

$$4p^2-232p+3300=0$$

$$p^2-58p+825=0$$

$D=(-58)^2-4\cdot 825=$$3364-3 300=64$

$p_1=\frac{-(-58)+\sqrt{64} }{2}=\dfrac{58+8}2=$$\dfrac{66}2=33$

$$p_2=\dfrac{58-8}2=\dfrac{50}2=25$$

$$25>33$$