1)5 x^2 +14x-3=0 36x^2-25=0 4x^2=16x (x-3)^2-2(x-3)-15=0 2)одна из двух натуральных чисел...

1.1) 5x² + 14x - 3 = 0

D = 196 + 60 = 256

x₁ = $\frac{-14+16}{10} = \frac{2}{10} = \frac{1}{5}$

x₂ = $\frac{-14-16}{10} = \frac{-30}{10} = -3$

1.2) 36x² - 25 = 0

36x² = 25

x² = $\frac{25}{36}$

x = $\frac{5}{6}$

1.3. 4x² = 16x

4x² - 16x = 0

4x (x - 4) = 0

x₁ = 0

x₂ = 4

1.4. (x-3)²-2(x-3)-15=0

x² - 6x + 9 - 2x + 6 -15 = 0

x² - 6x -2x + 9 + 6 - 15 = 0

x² - 8x = 0

x(x-8) = 0

x₁ = 0

x₂ = 8

2) Обыкновенная система:

x - y = 3
xy = 180

x = y + 3

y(y + 3) = 180

y² + 3y - 180 = 0

D = 9 + 720 = 729

y₁ = $\frac{-3+27}{2} = \frac{24}{2} = 12$

y₂ = $\frac{-3-27}{2} = \frac{-30}{2} = -15$ (не подходит)

x = 12+3 = 15

Удачи!