What are the solutions to the equation 5x^2+320=0 ?

The solutions to the equation 5x^2+320=0 are x=8i,x=-8i

Find the zeros of 5x^2+320=0

The zeros of 5x^2+320=0 are x=8i,x=-8i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + 320 = 0

x = 8 i, x = - 8 i

Solution steps

5 x^{2} + 320 = 0

Move

320

to the right side

5 x^{2} = - 320

Divide both sides by

5

x^{2} = - 64

For

x^{2} = f (a)

the solutions are

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

x = - 64, x = - - 64

- 64 = 8 i

- - 64 = - 8 i

x = 8 i, x = - 8 i

∣ - 3 w + 6 ∣ - 10 = 2

14 m^{2} - 31 m - 10

5 x^{2} - 7 x - 6 = 0

9 \leq \frac{11 x - 18}{9} < 20

What are the solutions to the equation 5x^2+320=0 ?
The solutions to the equation 5x^2+320=0 are x=8i,x=-8i
Find the zeros of 5x^2+320=0
The zeros of 5x^2+320=0 are x=8i,x=-8i