What are the solutions to the equation 240x^3+720x^2-600x=0 ?

The solutions to the equation 240x^3+720x^2-600x=0 are x=0,x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2

Find the zeros of 240x^3+720x^2-600x=0

The zeros of 240x^3+720x^2-600x=0 are x=0,x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2

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240 x^{3} + 720 x^{2} - 600 x = 0

x = 0, x = \frac{- 3 + 19}{2}, x = - \frac{3 + 19}{2}

Decimal Notation

x = 0, x = 0.67944 \dots, x = - 3.67944 \dots

Solution steps

240 x^{3} + 720 x^{2} - 600 x = 0

Factor

240 x^{3} + 720 x^{2} - 600 x : 120 x (2 x^{2} + 6 x - 5)

120 x (2 x^{2} + 6 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x^{2} + 6 x - 5 = 0

Solve

2 x^{2} + 6 x - 5 = 0 : x = \frac{- 3 + 19}{2}, x = - \frac{3 + 19}{2}

The solutions are

x = 0, x = \frac{- 3 + 19}{2}, x = - \frac{3 + 19}{2}

z^{3} = (3 + i)

x^{3} - 3 x^{2} - 10 x = - 24

8 + 4 x^{9} - 8 = 100

m^{3} = 1

r^{3} - 3 r^{2} + 3 r - 1

What are the solutions to the equation 240x^3+720x^2-600x=0 ?
The solutions to the equation 240x^3+720x^2-600x=0 are x=0,x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2
Find the zeros of 240x^3+720x^2-600x=0
The zeros of 240x^3+720x^2-600x=0 are x=0,x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2