What are the solutions to the equation 3x^2+12x+8=0 ?

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

Find the zeros of 3x^2+12x+8=0

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

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

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

Decimal Notation

x = - 0.84529 \dots, x = - 3.15470 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 8 = 43

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

Separate the solutions

x_{1} = \frac{- 12 + 4 3}{2 \cdot 3}, x_{2} = \frac{- 12 - 4 3}{2 \cdot 3}

x = \frac{- 12 + 4 3}{2 \cdot 3} : \frac{2 ( 3 - 3 )}{3}

x = \frac{- 12 - 4 3}{2 \cdot 3} : - \frac{2 ( 3 + 3 )}{3}

The solutions to the quadratic equation are:

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

3 z + 18 = 2 z + 6

x^{2} + x - 2 \leq 0

(8 x)^{2}

x = 0.25 \cdot (1 0^{4} x + 0.8)^{2}

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