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

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

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

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

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3 x (x + 2) = 1

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

Decimal Notation

x = 0.15470 \dots, x = - 2.15470 \dots

Solution steps

3 x (x + 2) = 1

Expand

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

3 x^{2} + 6 x = 1

Move

1

to the left side

3 x^{2} + 6 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 1) = 43

x_{1, 2} = \frac{- 6 \pm 4 3}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 4 3}{2 \cdot 3}, x_{2} = \frac{- 6 - 4 3}{2 \cdot 3}

x = \frac{- 6 + 4 3}{2 \cdot 3} : \frac{- 3 + 2 3}{3}

x = \frac{- 6 - 4 3}{2 \cdot 3} : - \frac{3 + 2}{3}

The solutions to the quadratic equation are:

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

\frac{x - 4}{3} + 1 > x + 5

- \frac{27}{2} (\frac{1}{100}) + \frac{27}{2}

- 1 x + 115 < 243

3 t^{2} - 12 t + 9

2 \cdot 8 \cdot 4

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