What are the solutions to the equation 0=3x^2-4x-2 ?

The solutions to the equation 0=3x^2-4x-2 are x=(2+sqrt(10))/3 ,x=(2-sqrt(10))/3

Find the zeros of 0=3x^2-4x-2

The zeros of 0=3x^2-4x-2 are x=(2+sqrt(10))/3 ,x=(2-sqrt(10))/3

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0 = 3 x^{2} - 4 x - 2

x = \frac{2 + 10}{3}, x = \frac{2 - 10}{3}

Decimal Notation

x = 1.72075 \dots, x = - 0.38742 \dots

Solution steps

0 = 3 x^{2} - 4 x - 2

Switch sides

3 x^{2} - 4 x - 2 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 3 (- 2) = 210

x_{1, 2} = \frac{- ( - 4 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 10}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 10}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 10}{2 \cdot 3} : \frac{2 + 10}{3}

x = \frac{- ( - 4 ) - 2 10}{2 \cdot 3} : \frac{2 - 10}{3}

The solutions to the quadratic equation are:

x = \frac{2 + 10}{3}, x = \frac{2 - 10}{3}

4 x^{2} = 10 x

4 x^{2} = 13 x

x, x^{2} + 2 x y + 4 = 2 x y

\frac{x ^{2} + 1}{2} = 2 x - 1

5 - x = 16 - 16 x + 4 x^{2}

What are the solutions to the equation 0=3x^2-4x-2 ?
The solutions to the equation 0=3x^2-4x-2 are x=(2+sqrt(10))/3 ,x=(2-sqrt(10))/3
Find the zeros of 0=3x^2-4x-2
The zeros of 0=3x^2-4x-2 are x=(2+sqrt(10))/3 ,x=(2-sqrt(10))/3