What are the solutions to the equation 6x^2+12x+3=0 ?

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

Find the zeros of 6x^2+12x+3=0

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

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

x = \frac{- 2 + 2}{2}, x = - \frac{2 + 1}{2}

Decimal Notation

x = - 0.29289 \dots, x = - 1.70710 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

1 2^{2} - 4 \cdot 6 \cdot 3 = 62

x_{1, 2} = \frac{- 12 \pm 6 2}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 12 + 6 2}{2 \cdot 6}, x_{2} = \frac{- 12 - 6 2}{2 \cdot 6}

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

x = \frac{- 12 - 6 2}{2 \cdot 6} : - \frac{2 + 1}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 2}{2}, x = - \frac{2 + 1}{2}

\frac{\frac{4}{x + h} - \frac{4}{x}}{h}

5 x - 4 \geq 6

16 x^{2} - 2 x y

52 x + 52 = 52 x - 78

9 - c = - 13

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