What are the solutions to the equation 7p^2+198=99 ?

The solutions to the equation 7p^2+198=99 are p=(3sqrt(11))/(sqrt(7))i,p=-(3sqrt(11))/(sqrt(7))i

Find the zeros of 7p^2+198=99

The zeros of 7p^2+198=99 are p=(3sqrt(11))/(sqrt(7))i,p=-(3sqrt(11))/(sqrt(7))i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

7 p^{2} + 198 = 99

p = \frac{3 11}{7} i, p = - \frac{3 11}{7} i

Solution steps

7 p^{2} + 198 = 99

Move

198

to the right side

7 p^{2} = - 99

Divide both sides by

7

p^{2} = - \frac{99}{7}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

p = - \frac{99}{7}, p = - - \frac{99}{7}

- \frac{99}{7} = \frac{3 11}{7} i

- - \frac{99}{7} = - \frac{3 11}{7} i

p = \frac{3 11}{7} i, p = - \frac{3 11}{7} i

y^{2} = 12 y

3 5^{2}

- \frac{2}{12} + (- \frac{4}{6}) + \frac{5}{3}

25 - 16 x^{2} \geq 0

\frac{1}{2} + \frac{12}{x ^{2}} > \frac{5}{x}

What are the solutions to the equation 7p^2+198=99 ?
The solutions to the equation 7p^2+198=99 are p=(3sqrt(11))/(sqrt(7))i,p=-(3sqrt(11))/(sqrt(7))i
Find the zeros of 7p^2+198=99
The zeros of 7p^2+198=99 are p=(3sqrt(11))/(sqrt(7))i,p=-(3sqrt(11))/(sqrt(7))i