What are the solutions to the equation 6p^2=-31p-5 ?

The solutions to the equation 6p^2=-31p-5 are p=-1/6 ,p=-5

Find the zeros of 6p^2=-31p-5

The zeros of 6p^2=-31p-5 are p=-1/6 ,p=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 p^{2} = - 31 p - 5

p = - \frac{1}{6}, p = - 5

Decimal Notation

p = - 0.16666 \dots, p = - 5

Solution steps

6 p^{2} = - 31 p - 5

Move

5

to the left side

6 p^{2} + 5 = - 31 p

Move

31 p

to the left side

6 p^{2} + 5 + 31 p = 0

Write in the standard form

a x^{2} + b x + c = 0

6 p^{2} + 31 p + 5 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

3 1^{2} - 4 \cdot 6 \cdot 5 = 29

p_{1, 2} = \frac{- 31 \pm 29}{2 \cdot 6}

Separate the solutions

p_{1} = \frac{- 31 + 29}{2 \cdot 6}, p_{2} = \frac{- 31 - 29}{2 \cdot 6}

p = \frac{- 31 + 29}{2 \cdot 6} : - \frac{1}{6}

p = \frac{- 31 - 29}{2 \cdot 6} : - 5

The solutions to the quadratic equation are:

p = - \frac{1}{6}, p = - 5

0 = 32 - 50 x^{2}

x^{12} - 64

\frac{9 x}{2} - \frac{x}{6} = 2

y (y - 1) - (y - 5)^{2} = 2

\frac{1}{3} (\frac{x - 1}{2}) < x - 3

What are the solutions to the equation 6p^2=-31p-5 ?
The solutions to the equation 6p^2=-31p-5 are p=-1/6 ,p=-5
Find the zeros of 6p^2=-31p-5
The zeros of 6p^2=-31p-5 are p=-1/6 ,p=-5