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 ( - 3 ) \cdot 8}{2 ( - 3 )}

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

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

Separate the solutions

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

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

x = \frac{- 12 - 4 15}{2 ( - 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}

x^{2} - 4 x + 14 = 13

3 x^{2} - 2 x + 18 = 0

2 x^{2} + 7 x - 13 = - 10

24 = 4 + 37 t - 16 t^{2}

x, 4 x^{2} y + 2 y^{3} = 1 + 6 x y^{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(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