What are the solutions to the equation 12x^2+24x+8=0 ?

The solutions to the equation 12x^2+24x+8=0 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))

Find the zeros of 12x^2+24x+8=0

The zeros of 12x^2+24x+8=0 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))

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12 x^{2} + 24 x + 8 = 0

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

Decimal Notation

x = - 0.42264 \dots, x = - 1.57735 \dots

Solution steps

12 x^{2} + 24 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 12 \cdot 8}{2 \cdot 12}

2 4^{2} - 4 \cdot 12 \cdot 8 = 83

x_{1, 2} = \frac{- 24 \pm 8 3}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 24 + 8 3}{2 \cdot 12}, x_{2} = \frac{- 24 - 8 3}{2 \cdot 12}

x = \frac{- 24 + 8 3}{2 \cdot 12} : \frac{- 3 + 3}{3}

x = \frac{- 24 - 8 3}{2 \cdot 12} : - \frac{3 + 1}{3}

The solutions to the quadratic equation are:

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

(6 - x) (x + 1) = 0

0 = 210 + 205.06 x - 4.9 x^{2}

(x - 1)^{2} - 64 = 0

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

x^{2} + \frac{1}{3} x + 2 = 0

What are the solutions to the equation 12x^2+24x+8=0 ?
The solutions to the equation 12x^2+24x+8=0 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))
Find the zeros of 12x^2+24x+8=0
The zeros of 12x^2+24x+8=0 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))