What are the solutions to the equation 9p^2+6p+1=12 ?

The solutions to the equation 9p^2+6p+1=12 are p=(-1+2sqrt(3))/3 ,p=-(1+2sqrt(3))/3

Find the zeros of 9p^2+6p+1=12

The zeros of 9p^2+6p+1=12 are p=(-1+2sqrt(3))/3 ,p=-(1+2sqrt(3))/3

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9 p^{2} + 6 p + 1 = 12

p = \frac{- 1 + 2 3}{3}, p = - \frac{1 + 2 3}{3}

Decimal

p = 0.82136 \dots, p = - 1.48803 \dots

Solution steps

9 p^{2} + 6 p + 1 = 12

Move

12

to the left side

9 p^{2} + 6 p - 11 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 11 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 11) = 123

p_{1, 2} = \frac{- 6 \pm 12 3}{2 \cdot 9}

Separate the solutions

p_{1} = \frac{- 6 + 12 3}{2 \cdot 9}, p_{2} = \frac{- 6 - 12 3}{2 \cdot 9}

p = \frac{- 6 + 12 3}{2 \cdot 9} : \frac{- 1 + 2 3}{3}

p = \frac{- 6 - 12 3}{2 \cdot 9} : - \frac{1 + 2 3}{3}

The solutions to the quadratic equation are:

p = \frac{- 1 + 2 3}{3}, p = - \frac{1 + 2 3}{3}

x^{2} = 9 x + 2

(14 x + 21) (1.8 - 0.3 x) = 0

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

x^{2} + 3 x + 8 = 5

a y^{2}

What are the solutions to the equation 9p^2+6p+1=12 ?
The solutions to the equation 9p^2+6p+1=12 are p=(-1+2sqrt(3))/3 ,p=-(1+2sqrt(3))/3
Find the zeros of 9p^2+6p+1=12
The zeros of 9p^2+6p+1=12 are p=(-1+2sqrt(3))/3 ,p=-(1+2sqrt(3))/3