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

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

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

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

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

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

Decimal Notation

p = 0.30940 \dots, p = - 4.30940 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{2} - 4 x - 1 = 2 x^{2} - 5

4 x^{2} + x - 5 = - 6 x

4 x^{2} + 3 x = x (2 x - 2)

(x - 6) (2 x + 1) = 0

12 x = - 4 x^{2}

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