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

The solutions to the equation p^2+np-4=0 are p=(-n+sqrt(n^2+16))/2 ,p=(-n-sqrt(n^2+16))/2

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

The zeros of p^2+np-4=0 are p=(-n+sqrt(n^2+16))/2 ,p=(-n-sqrt(n^2+16))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

p = \frac{- n + n ^{2} + 16}{2}, p = \frac{- n - n ^{2} + 16}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

n^{2} - 4 \cdot 1 \cdot (- 4) : n^{2} + 16

p_{1, 2} = \frac{- n \pm n ^{2} + 16}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- n + n ^{2} + 16}{2 \cdot 1}, p_{2} = \frac{- n - n ^{2} + 16}{2 \cdot 1}

p = \frac{- n + n ^{2} + 16}{2 \cdot 1} : \frac{- n + n ^{2} + 16}{2}

p = \frac{- n - n ^{2} + 16}{2 \cdot 1} : \frac{- n - n ^{2} + 16}{2}

The solutions to the quadratic equation are:

p = \frac{- n + n ^{2} + 16}{2}, p = \frac{- n - n ^{2} + 16}{2}

(x + 19)^{2} = - 4

r, a = 4 π r^{2}

x^{2} + 4 x + 12

z^{2} - 0.3063 z + 0.4317 = 0

2 x^{2} + 2 x - 10 = 0

What are the solutions to the equation p^2+np-4=0 ?
The solutions to the equation p^2+np-4=0 are p=(-n+sqrt(n^2+16))/2 ,p=(-n-sqrt(n^2+16))/2
Find the zeros of p^2+np-4=0
The zeros of p^2+np-4=0 are p=(-n+sqrt(n^2+16))/2 ,p=(-n-sqrt(n^2+16))/2