What are the solutions to the equation p^2+7p+2=0 ?

The solutions to the equation p^2+7p+2=0 are p=(-7+sqrt(41))/2 ,p=(-7-sqrt(41))/2

Find the zeros of p^2+7p+2=0

The zeros of p^2+7p+2=0 are p=(-7+sqrt(41))/2 ,p=(-7-sqrt(41))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

p^{2} + 7 p + 2 = 0

p = \frac{- 7 + 41}{2}, p = \frac{- 7 - 41}{2}

Decimal Notation

p = - 0.29843 \dots, p = - 6.70156 \dots

Solution steps

p^{2} + 7 p + 2 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 2 = 41

p_{1, 2} = \frac{- 7 \pm 41}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 7 + 41}{2 \cdot 1}, p_{2} = \frac{- 7 - 41}{2 \cdot 1}

p = \frac{- 7 + 41}{2 \cdot 1} : \frac{- 7 + 41}{2}

p = \frac{- 7 - 41}{2 \cdot 1} : \frac{- 7 - 41}{2}

The solutions to the quadratic equation are:

p = \frac{- 7 + 41}{2}, p = \frac{- 7 - 41}{2}

3 x^{2} - 17 x = - 10

(x - 5)^{2} = 18

15 x^{2} + 840 x - 3600 = 0

3 b^{2} - 31 b + 56 = 0

6^{2} = x (13 - x)

What are the solutions to the equation p^2+7p+2=0 ?
The solutions to the equation p^2+7p+2=0 are p=(-7+sqrt(41))/2 ,p=(-7-sqrt(41))/2
Find the zeros of p^2+7p+2=0
The zeros of p^2+7p+2=0 are p=(-7+sqrt(41))/2 ,p=(-7-sqrt(41))/2