What are the solutions to the equation 3p^2+p-9=0 ?

The solutions to the equation 3p^2+p-9=0 are p=(-1+sqrt(109))/6 ,p=(-1-sqrt(109))/6

Find the zeros of 3p^2+p-9=0

The zeros of 3p^2+p-9=0 are p=(-1+sqrt(109))/6 ,p=(-1-sqrt(109))/6

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3 p^{2} + p - 9 = 0

p = \frac{- 1 + 109}{6}, p = \frac{- 1 - 109}{6}

Decimal Notation

p = 1.57338 \dots, p = - 1.90671 \dots

Solution steps

3 p^{2} + p - 9 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 9) = 109

p_{1, 2} = \frac{- 1 \pm 109}{2 \cdot 3}

Separate the solutions

p_{1} = \frac{- 1 + 109}{2 \cdot 3}, p_{2} = \frac{- 1 - 109}{2 \cdot 3}

p = \frac{- 1 + 109}{2 \cdot 3} : \frac{- 1 + 109}{6}

p = \frac{- 1 - 109}{2 \cdot 3} : \frac{- 1 - 109}{6}

The solutions to the quadratic equation are:

p = \frac{- 1 + 109}{6}, p = \frac{- 1 - 109}{6}

4 x^{2} - 4 x + 1

\frac{e ^{4} - e ^{- 4}}{2}

4 x^{2} - 9 x + 4 = 0

x

\frac{( x ^{2} - 1 )}{x ^{2} + 5 x + 4} \leq 0

What are the solutions to the equation 3p^2+p-9=0 ?
The solutions to the equation 3p^2+p-9=0 are p=(-1+sqrt(109))/6 ,p=(-1-sqrt(109))/6
Find the zeros of 3p^2+p-9=0
The zeros of 3p^2+p-9=0 are p=(-1+sqrt(109))/6 ,p=(-1-sqrt(109))/6