What are the solutions to the equation 6x^2+4x-12=0 ?

The solutions to the equation 6x^2+4x-12=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3

Find the zeros of 6x^2+4x-12=0

The zeros of 6x^2+4x-12=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3

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6 x^{2} + 4 x - 12 = 0

x = \frac{- 1 + 19}{3}, x = - \frac{1 + 19}{3}

Decimal Notation

x = 1.11963 \dots, x = - 1.78629 \dots

Solution steps

6 x^{2} + 4 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 6 ( - 12 )}{2 \cdot 6}

4^{2} - 4 \cdot 6 (- 12) = 419

x_{1, 2} = \frac{- 4 \pm 4 19}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 4 + 4 19}{2 \cdot 6}, x_{2} = \frac{- 4 - 4 19}{2 \cdot 6}

x = \frac{- 4 + 4 19}{2 \cdot 6} : \frac{- 1 + 19}{3}

x = \frac{- 4 - 4 19}{2 \cdot 6} : - \frac{1 + 19}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 19}{3}, x = - \frac{1 + 19}{3}

e = \frac{1}{2} m v^{2}

x, \frac{x ^{2}}{4} + y^{2} + \frac{z ^{2}}{9} = 3

12 x^{2} = 108

x, x^{2} - k x + 169 = 0

4 x^{2} + 5 = - 9 x

What are the solutions to the equation 6x^2+4x-12=0 ?
The solutions to the equation 6x^2+4x-12=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3
Find the zeros of 6x^2+4x-12=0
The zeros of 6x^2+4x-12=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3