What are the solutions to the equation 4p^2+p-6=0 ?

The solutions to the equation 4p^2+p-6=0 are p=(-1+sqrt(97))/8 ,p=(-1-sqrt(97))/8

Find the zeros of 4p^2+p-6=0

The zeros of 4p^2+p-6=0 are p=(-1+sqrt(97))/8 ,p=(-1-sqrt(97))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 p^{2} + p - 6 = 0

p = \frac{- 1 + 97}{8}, p = \frac{- 1 - 97}{8}

Decimal Notation

p = 1.10610 \dots, p = - 1.35610 \dots

Solution steps

4 p^{2} + p - 6 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 6) = 97

p_{1, 2} = \frac{- 1 \pm 97}{2 \cdot 4}

Separate the solutions

p_{1} = \frac{- 1 + 97}{2 \cdot 4}, p_{2} = \frac{- 1 - 97}{2 \cdot 4}

p = \frac{- 1 + 97}{2 \cdot 4} : \frac{- 1 + 97}{8}

p = \frac{- 1 - 97}{2 \cdot 4} : \frac{- 1 - 97}{8}

The solutions to the quadratic equation are:

p = \frac{- 1 + 97}{8}, p = \frac{- 1 - 97}{8}

3 x^{2} + 16 x + 1 = 9 x - 3

- 6 x^{2} = - 24

9 k^{2} - 18 k + 5 = 0

x, m x^{2} - m x + 3 = 0

\frac{2}{3} x^{2} = 6

What are the solutions to the equation 4p^2+p-6=0 ?
The solutions to the equation 4p^2+p-6=0 are p=(-1+sqrt(97))/8 ,p=(-1-sqrt(97))/8
Find the zeros of 4p^2+p-6=0
The zeros of 4p^2+p-6=0 are p=(-1+sqrt(97))/8 ,p=(-1-sqrt(97))/8