What are the solutions to the equation 2x^3+3x^2-7x-4=0 ?

The solutions to the equation 2x^3+3x^2-7x-4=0 are x=-1/2 ,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

Find the zeros of 2x^3+3x^2-7x-4=0

The zeros of 2x^3+3x^2-7x-4=0 are x=-1/2 ,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

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2 x^{3} + 3 x^{2} - 7 x - 4 = 0

x = - \frac{1}{2}, x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

Decimal Notation

x = - 0.5, x = 1.56155 \dots, x = - 2.56155 \dots

Solution steps

2 x^{3} + 3 x^{2} - 7 x - 4 = 0

Factor

2 x^{3} + 3 x^{2} - 7 x - 4 : (2 x + 1) (x^{2} + x - 4)

(2 x + 1) (x^{2} + x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0 or x^{2} + x - 4 = 0

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x^{2} + x - 4 = 0 : x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

The solutions are

x = - \frac{1}{2}, x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

x^{8} = - 3

20 x^{3} - 18 x = 0

x^{3} - 3 x^{2} - 10 x = 0

x^{2} - 4 x = x^{3} - 6 x^{2} + 8 x

s^{3} + s^{2} + 4 s + 4

What are the solutions to the equation 2x^3+3x^2-7x-4=0 ?
The solutions to the equation 2x^3+3x^2-7x-4=0 are x=-1/2 ,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2
Find the zeros of 2x^3+3x^2-7x-4=0
The zeros of 2x^3+3x^2-7x-4=0 are x=-1/2 ,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2