What is f(x)=0,f(x)=x^3-3x^2+4 ?

The solution to f(x)=0,f(x)=x^3-3x^2+4 is

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f(x)=0,f(x)=x^3-3x^2+4

f (x) = 0, f (x) = x^{3} - 3 x^{2} + 4

(x = - 1, x = 2, f = 0 f = 0)

Solution steps

[f = 0 f = x^{3} - 3 x^{2} + 4]

Plug the solutions

f = 0

into

f = x^{3} - 3 x^{2} + 4

For

f = x^{3} - 3 x^{2} + 4

, subsitute

f

with

0 : x = - 1, x = 2

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

f = 0, f = x^{3} - 3 x^{2} + 4

are

(x = - 1, x = 2, f = 0 f = 0)

d = 300 m, t = 3 h

c = (a + b) + c, a = 6

2 x + 3 y = 9, (a + b) x + (2 a - b) y = 3 (a + b + 1)

2 x + y = 2, x y x + 2 y = 4 x y

6 x^{2} - 40 x + 24 = 0, 6 y^{2} + 100 y - 34 = 0