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(33))/3 ,x=-(6+sqrt(33))/3

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

The zeros of 3x^2+12x+1=0 are x=(-6+sqrt(33))/3 ,x=-(6+sqrt(33))/3

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

x = \frac{- 6 + 33}{3}, x = - \frac{6 + 33}{3}

Decimal Notation

x = - 0.08514 \dots, x = - 3.91485 \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 \cdot 1}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 1 = 233

x_{1, 2} = \frac{- 12 \pm 2 33}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 2 33}{2 \cdot 3}, x_{2} = \frac{- 12 - 2 33}{2 \cdot 3}

x = \frac{- 12 + 2 33}{2 \cdot 3} : \frac{- 6 + 33}{3}

x = \frac{- 12 - 2 33}{2 \cdot 3} : - \frac{6 + 33}{3}

The solutions to the quadratic equation are:

x = \frac{- 6 + 33}{3}, x = - \frac{6 + 33}{3}

∣ x + 3 ∣ = 8

4 \leq \frac{u}{- 3}

1.5 e^{1.5 \cdot \frac{2}{3}} ((\frac{2}{3})^{2} - \frac{2}{3})

x^{2} y^{3} + 2 x^{3} y^{4} + x y y

9 x^{2} = - 324

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(33))/3 ,x=-(6+sqrt(33))/3
Find the zeros of 3x^2+12x+1=0
The zeros of 3x^2+12x+1=0 are x=(-6+sqrt(33))/3 ,x=-(6+sqrt(33))/3