д) у² = 52у - 576; e) 15y² - 30 = 22y + 7; ж) 25р² = 10p – 1; з) 299x² + 100x = 500 - 101x;

д) $$y^2 = 52y - 576$$

$$y^2 - 52y + 576 = 0$$

$$D = (-52)^2 - 4 \cdot 1 \cdot 576 = 2704 - 2304 = 400$$

$$y_1 = \frac{52 + \sqrt{400}}{2 \cdot 1} = \frac{52 + 20}{2} = \frac{72}{2} = 36$$

$$y_2 = \frac{52 - \sqrt{400}}{2 \cdot 1} = \frac{52 - 20}{2} = \frac{32}{2} = 16$$

Ответ: $$y_1 = 36$$, $$y_2 = 16$$

e) $$15y^2 - 30 = 22y + 7$$

$$15y^2 - 22y - 37 = 0$$

$$D = (-22)^2 - 4 \cdot 15 \cdot (-37) = 484 + 2220 = 2704$$

$$y_1 = \frac{22 + \sqrt{2704}}{2 \cdot 15} = \frac{22 + 52}{30} = \frac{74}{30} = \frac{37}{15}$$

$$y_2 = \frac{22 - \sqrt{2704}}{2 \cdot 15} = \frac{22 - 52}{30} = \frac{-30}{30} = -1$$

Ответ: $$y_1 = \frac{37}{15}$$, $$y_2 = -1$$

ж) $$25p^2 = 10p - 1$$

$$25p^2 - 10p + 1 = 0$$

$$D = (-10)^2 - 4 \cdot 25 \cdot 1 = 100 - 100 = 0$$

$$p = \frac{10 + \sqrt{0}}{2 \cdot 25} = \frac{10}{50} = \frac{1}{5} = 0.2$$

Ответ: $$p = 0.2$$

з) $$299x^2 + 100x = 500 - 101x$$

$$299x^2 + 201x - 500 = 0$$

$$D = 201^2 - 4 \cdot 299 \cdot (-500) = 40401 + 598000 = 638401$$

$$x_1 = \frac{-201 + \sqrt{638401}}{2 \cdot 299} = \frac{-201 + 799}{598} = \frac{598}{598} = 1$$

$$x_2 = \frac{-201 - \sqrt{638401}}{2 \cdot 299} = \frac{-201 - 799}{598} = \frac{-1000}{598} = \frac{-500}{299}$$

Ответ: $$x_1 = 1$$, $$x_2 = \frac{-500}{299}$$