What are the solutions to the equation 2x^3+6x^2+x=0 ?

The solutions to the equation 2x^3+6x^2+x=0 are x=0,x=(-3+sqrt(7))/2 ,x=-(3+sqrt(7))/2

Find the zeros of 2x^3+6x^2+x=0

The zeros of 2x^3+6x^2+x=0 are x=0,x=(-3+sqrt(7))/2 ,x=-(3+sqrt(7))/2

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2 x^{3} + 6 x^{2} + x = 0

x = 0, x = \frac{- 3 + 7}{2}, x = - \frac{3 + 7}{2}

Decimal Notation

x = 0, x = - 0.17712 \dots, x = - 2.82287 \dots

Solution steps

2 x^{3} + 6 x^{2} + x = 0

Factor

2 x^{3} + 6 x^{2} + x : x (2 x^{2} + 6 x + 1)

x (2 x^{2} + 6 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x^{2} + 6 x + 1 = 0

Solve

2 x^{2} + 6 x + 1 = 0 : x = \frac{- 3 + 7}{2}, x = - \frac{3 + 7}{2}

The solutions are

x = 0, x = \frac{- 3 + 7}{2}, x = - \frac{3 + 7}{2}

5 x \cdot x \cdot 3 x = 1875

(6 x + 10) (7 x - 2) (8 x + 4) = 180

3 x^{4} + 3 x^{3} - 17 x^{2} + x - 6 = 0

t^{3} = 512

x^{3} - 5 x^{2} + 5 x + 7 = 7

What are the solutions to the equation 2x^3+6x^2+x=0 ?
The solutions to the equation 2x^3+6x^2+x=0 are x=0,x=(-3+sqrt(7))/2 ,x=-(3+sqrt(7))/2
Find the zeros of 2x^3+6x^2+x=0
The zeros of 2x^3+6x^2+x=0 are x=0,x=(-3+sqrt(7))/2 ,x=-(3+sqrt(7))/2