What are the solutions to the equation -3x^2+12x+4=0 ?

The solutions to the equation -3x^2+12x+4=0 are x=-(2(2sqrt(3)-3))/3 ,x=(2(3+2sqrt(3)))/3

Find the zeros of -3x^2+12x+4=0

The zeros of -3x^2+12x+4=0 are x=-(2(2sqrt(3)-3))/3 ,x=(2(3+2sqrt(3)))/3

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- 3 x^{2} + 12 x + 4 = 0

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

Decimal Notation

x = - 0.30940 \dots, x = 4.30940 \dots

Solution steps

- 3 x^{2} + 12 x + 4 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 (- 3) \cdot 4 = 83

x_{1, 2} = \frac{- 12 \pm 8 3}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 12 + 8 3}{2 ( - 3 )}, x_{2} = \frac{- 12 - 8 3}{2 ( - 3 )}

x = \frac{- 12 + 8 3}{2 ( - 3 )} : - \frac{2 ( 2 3 - 3 )}{3}

x = \frac{- 12 - 8 3}{2 ( - 3 )} : \frac{2 ( 3 + 2 3 )}{3}

The solutions to the quadratic equation are:

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

5 x^{2} - 3 + 2 x^{2} = 4

2 x^{2} + 12 x = 4

6 x^{2} - 7 x - 3 = 0

4 (x + 1)^{2} = 9

4 x^{2} + 268 = - 72 x

What are the solutions to the equation -3x^2+12x+4=0 ?
The solutions to the equation -3x^2+12x+4=0 are x=-(2(2sqrt(3)-3))/3 ,x=(2(3+2sqrt(3)))/3
Find the zeros of -3x^2+12x+4=0
The zeros of -3x^2+12x+4=0 are x=-(2(2sqrt(3)-3))/3 ,x=(2(3+2sqrt(3)))/3