Please use the root formula of cubic equation of one variable to solve the following equation: 2x ^ 3-x ^ 2-3x = 0

Please use the root formula of cubic equation of one variable to solve the following equation: 2x ^ 3-x ^ 2-3x = 0

2x^3-x^2-3x=0
x(2x² -x-3) = 0
x(2x-3) (x+1)= 0
X = 0 or x = 3 / 2 or x = - 1

The middle term of the equal ratio of the two parts of equation 3x ^ 2-7x 1 = 0 is

The product of two parts of the equation 3x ^ 2-7x + 1 = 0 = C / a = 1 / 3, and the equal proportion middle term of two parts = √ (x1 * x2) = √ (C / a) = √ (1 / 3) = √ 3 / 3
The product of two parts of equation 3x ^ 2-7x-1 = 0 = - 1 / 3, and there is no equal middle term

If the function f (x) = 1 / 3x ^ 3-4x + 4, the number of zeros of the equation f (x) = K in the interval [- 3,5] is discussed

F '(x) = x ^ 2-4 when x ≤ - 2, f' (x) ≥ 0, the function monotonically increases; when - 2 ≤ x ≤ 2, f '(x) ≤ 0, the function monotonically decreases; when x ≥ 2, f' (x) ≥ 0, the function monotonically increases ｜ f (x) takes the maximum value when x = - 2, f (- 2) = 28 / 3; when x = 2, takes the minimum value, f (2) = - 4 / 3, f (- 3) = 7, f (5) = 77 / 3, roughly draws f (x) in [...]