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

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

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

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

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

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

Decimal Notation

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

Solution steps

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}

7 - 1 [6 - 10 (y + 1)] > 2 {5 - 1 [2 - (y - 3)]}

0 = - 16 t^{2} + 95 t + 6

0 = 1.5 x^{2} - 4.5 x

\frac{16}{25} x^{2} y^{2} - \frac{36}{49} y^{4}

3 x = 42

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