What are the solutions to the equation 3x^2+12x-4=0 ?

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

Find the zeros of 3x^2+12x-4=0

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

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

x = \frac{2 ( 2 3 - 3 )}{3}, x = - \frac{2 ( 3 + 2 3 )}{3}

Decimal Notation

x = 0.30940 \dots, x = - 4.30940 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 4) = 83

x_{1, 2} = \frac{- 12 \pm 8 3}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 8 3}{2 \cdot 3}, x_{2} = \frac{- 12 - 8 3}{2 \cdot 3}

x = \frac{- 12 + 8 3}{2 \cdot 3} : \frac{2 ( 2 3 - 3 )}{3}

x = \frac{- 12 - 8 3}{2 \cdot 3} : - \frac{2 ( 3 + 2 3 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 2 3 - 3 )}{3}, x = - \frac{2 ( 3 + 2 3 )}{3}

2 x^{2} - x - 7 = 0

2 (x^{2} - 6) = - 2 x

2^{2} + 1^{2} = x^{2}

s^{2} + s + 10 = 0

x^{2} - 7 = - 6 x

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