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

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

Find the zeros of 3x^2-2x-2=0

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

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

x = \frac{1 + 7}{3}, x = \frac{1 - 7}{3}

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 3 (- 2) = 27

x_{1, 2} = \frac{- ( - 2 ) \pm 2 7}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 7}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 7}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 7}{2 \cdot 3} : \frac{1 + 7}{3}

x = \frac{- ( - 2 ) - 2 7}{2 \cdot 3} : \frac{1 - 7}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 7}{3}, x = \frac{1 - 7}{3}

- x^{2} - 2 \geq 0

(\frac{4}{3} - \frac{7}{9}) - (\frac{- 9}{15} - \frac{- 12}{5})

\frac{x ^{- 6}}{x ^{4}}

x - 5 x + 2 < 8

(- \frac{1}{3}) + (- \frac{3}{4})

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