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

The solutions to the equation 3x^2+2x-4=0 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3

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

The zeros of 3x^2+2x-4=0 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3

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

x = \frac{- 1 + 13}{3}, x = - \frac{1 + 13}{3}

Decimal Notation

x = 0.86851 \dots, x = - 1.53518 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 3 (- 4) = 213

x_{1, 2} = \frac{- 2 \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 13}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 13}{2 \cdot 3}

x = \frac{- 2 + 2 13}{2 \cdot 3} : \frac{- 1 + 13}{3}

x = \frac{- 2 - 2 13}{2 \cdot 3} : - \frac{1 + 13}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 13}{3}, x = - \frac{1 + 13}{3}

lo g_{2} (\frac{1}{8}) = - 3

(3 x^{4} y)^{3}

m^{2} - 16 m + 63

x^{2} + 14 x - 16 = 0

5 t^{2} - 14 t + 6 = 0

What are the solutions to the equation 3x^2+2x-4=0 ?
The solutions to the equation 3x^2+2x-4=0 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3
Find the zeros of 3x^2+2x-4=0
The zeros of 3x^2+2x-4=0 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3