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(15)-3))/3 ,x=-(2(3+sqrt(15)))/3

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

The zeros of 3x^2+12x-8=0 are x=(2(sqrt(15)-3))/3 ,x=-(2(3+sqrt(15)))/3

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

x = \frac{2 ( 15 - 3 )}{3}, x = - \frac{2 ( 3 + 15 )}{3}

Decimal Notation

x = 0.58198 \dots, x = - 4.58198 \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 ( - 8 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 8) = 415

x_{1, 2} = \frac{- 12 \pm 4 15}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 4 15}{2 \cdot 3}, x_{2} = \frac{- 12 - 4 15}{2 \cdot 3}

x = \frac{- 12 + 4 15}{2 \cdot 3} : \frac{2 ( 15 - 3 )}{3}

x = \frac{- 12 - 4 15}{2 \cdot 3} : - \frac{2 ( 3 + 15 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 15 - 3 )}{3}, x = - \frac{2 ( 3 + 15 )}{3}

\frac{6}{7} x + 2 = 18

10 x^{2} - 24 x - 152 = 0

2 x^{2} - 10 x - 12

24 \cdot 3

(4 x - 2)^{2} = 17

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(15)-3))/3 ,x=-(2(3+sqrt(15)))/3
Find the zeros of 3x^2+12x-8=0
The zeros of 3x^2+12x-8=0 are x=(2(sqrt(15)-3))/3 ,x=-(2(3+sqrt(15)))/3