What are the solutions to the equation x(x+5)^2-x^3=1 ?

The solutions to the equation x(x+5)^2-x^3=1 are x=(-25+sqrt(665))/(20),x=(-25-sqrt(665))/(20)

Find the zeros of x(x+5)^2-x^3=1

The zeros of x(x+5)^2-x^3=1 are x=(-25+sqrt(665))/(20),x=(-25-sqrt(665))/(20)

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x (x + 5)^{2} - x^{3} = 1

x = \frac{- 25 + 665}{20}, x = \frac{- 25 - 665}{20}

Decimal Notation

x = 0.03937 \dots, x = - 2.53937 \dots

Solution steps

x (x + 5)^{2} - x^{3} = 1

Expand

x (x + 5)^{2} - x^{3} : 10 x^{2} + 25 x

10 x^{2} + 25 x = 1

Move

1

to the left side

10 x^{2} + 25 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

2 5^{2} - 4 \cdot 10 (- 1) = 665

x_{1, 2} = \frac{- 25 \pm 665}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 25 + 665}{2 \cdot 10}, x_{2} = \frac{- 25 - 665}{2 \cdot 10}

x = \frac{- 25 + 665}{2 \cdot 10} : \frac{- 25 + 665}{20}

x = \frac{- 25 - 665}{2 \cdot 10} : \frac{- 25 - 665}{20}

The solutions to the quadratic equation are:

x = \frac{- 25 + 665}{20}, x = \frac{- 25 - 665}{20}

x^{2} - 7 x - 638 = 0

x, 6 x y^{2} = - 12 x^{2} y

8 x^{2} + 6 x = 4

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

2 x^{2} + 3 x - 3 = 1

What are the solutions to the equation x(x+5)^2-x^3=1 ?
The solutions to the equation x(x+5)^2-x^3=1 are x=(-25+sqrt(665))/(20),x=(-25-sqrt(665))/(20)
Find the zeros of x(x+5)^2-x^3=1
The zeros of x(x+5)^2-x^3=1 are x=(-25+sqrt(665))/(20),x=(-25-sqrt(665))/(20)