What are the solutions to the equation p^2-12=4p ?

The solutions to the equation p^2-12=4p are p=6,p=-2

Find the zeros of p^2-12=4p

The zeros of p^2-12=4p are p=6,p=-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

p^{2} - 12 = 4 p

p = 6, p = - 2

Solution steps

p^{2} - 12 = 4 p

Move

4 p

to the left side

p^{2} - 12 - 4 p = 0

Write in the standard form

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

p^{2} - 4 p - 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

p_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, p_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

p = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

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

The solutions to the quadratic equation are:

p = 6, p = - 2

2 x^{2} - 7 x + 9 = 0

(3 x^{2})^{4}

- 19 = 15 t - (\frac{1}{2}) 9.8 (t^{2})

x^{2} = 19 x

2^{2} - 7 x + 3 = 0

What are the solutions to the equation p^2-12=4p ?
The solutions to the equation p^2-12=4p are p=6,p=-2
Find the zeros of p^2-12=4p
The zeros of p^2-12=4p are p=6,p=-2