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

The solutions to the equation (3x+1)^2=-2x are x=(-4+sqrt(7))/9 ,x=-(4+sqrt(7))/9

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

The zeros of (3x+1)^2=-2x are x=(-4+sqrt(7))/9 ,x=-(4+sqrt(7))/9

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

x = \frac{- 4 + 7}{9}, x = - \frac{4 + 7}{9}

Decimal Notation

x = - 0.15047 \dots, x = - 0.73841 \dots

Solution steps

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

Expand

(3 x + 1)^{2} : 9 x^{2} + 6 x + 1

9 x^{2} + 6 x + 1 = - 2 x

Move

2 x

to the left side

9 x^{2} + 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

8^{2} - 4 \cdot 9 \cdot 1 = 27

x_{1, 2} = \frac{- 8 \pm 2 7}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 8 + 2 7}{2 \cdot 9}, x_{2} = \frac{- 8 - 2 7}{2 \cdot 9}

x = \frac{- 8 + 2 7}{2 \cdot 9} : \frac{- 4 + 7}{9}

x = \frac{- 8 - 2 7}{2 \cdot 9} : - \frac{4 + 7}{9}

The solutions to the quadratic equation are:

x = \frac{- 4 + 7}{9}, x = - \frac{4 + 7}{9}

\frac{4 x}{x + 6} < x

x^{3} y - y^{3} x

(- 3 - 2 i) (- 4 + 2 i)

1 0^{- 0.6}

t^{2} - 16 = 6 t

What are the solutions to the equation (3x+1)^2=-2x ?
The solutions to the equation (3x+1)^2=-2x are x=(-4+sqrt(7))/9 ,x=-(4+sqrt(7))/9
Find the zeros of (3x+1)^2=-2x
The zeros of (3x+1)^2=-2x are x=(-4+sqrt(7))/9 ,x=-(4+sqrt(7))/9