Знайдить область визначення функции f(x)=3+2x-x²

f(x) = 3 + 2x - x²

0 = 3 + 2x - x²

3 + 2x - x² = 0

- x² + 2x + 3 = 0

x² - 2x - 3 = 0

x² + x - 3x - 3 = 0

x*(x + 1) - 3x - 3 = 0

x*(x + 1) - 3(x + 1) = 0

(x + 1)*(x - 3) = 0

x + 1 = 0

x - 3 = 0

x + 1 - 1 = 0 - 1

x = 0 - 1

x = -1

x = 3

x1 = -1, x2 = 3