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

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

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

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

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3 x^{2} - 2 x = 2

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

Decimal Notation

x = 1.21525 \dots, x = - 0.54858 \dots

Solution steps

3 x^{2} - 2 x = 2

Move

2

to the left side

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}

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

x + 5 = x^{2} - x + 2

3 x^{2} + 6 x - 7 = 0

m^{2} + 10 m + 16 = 0

- 2 x^{2} - 15 x - 4 = - 3 x^{2}

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