What are the solutions to the equation 3x^2+12x-1=0 ?

The solutions to the equation 3x^2+12x-1=0 are x=(-6+sqrt(39))/3 ,x=-(6+sqrt(39))/3

Find the zeros of 3x^2+12x-1=0

The zeros of 3x^2+12x-1=0 are x=(-6+sqrt(39))/3 ,x=-(6+sqrt(39))/3

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3 x^{2} + 12 x - 1 = 0

x = \frac{- 6 + 39}{3}, x = - \frac{6 + 39}{3}

Decimal Notation

x = 0.08166 \dots, x = - 4.08166 \dots

Solution steps

3 x^{2} + 12 x - 1 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 3 (- 1) = 239

x_{1, 2} = \frac{- 12 \pm 2 39}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 2 39}{2 \cdot 3}, x_{2} = \frac{- 12 - 2 39}{2 \cdot 3}

x = \frac{- 12 + 2 39}{2 \cdot 3} : \frac{- 6 + 39}{3}

x = \frac{- 12 - 2 39}{2 \cdot 3} : - \frac{6 + 39}{3}

The solutions to the quadratic equation are:

x = \frac{- 6 + 39}{3}, x = - \frac{6 + 39}{3}

x^{2} + 3 x - 1 = 0

3 x^{2} + 2 x - 7

8 x = - (16 x^{2} + 1)

3 x^{2} = 19 x - 20

(3 + x) \frac{2}{m} + 6 (3 - x) \frac{2}{m} = 5 (9 - x^{2}) \frac{1}{m}

What are the solutions to the equation 3x^2+12x-1=0 ?
The solutions to the equation 3x^2+12x-1=0 are x=(-6+sqrt(39))/3 ,x=-(6+sqrt(39))/3
Find the zeros of 3x^2+12x-1=0
The zeros of 3x^2+12x-1=0 are x=(-6+sqrt(39))/3 ,x=-(6+sqrt(39))/3