What are the solutions to the equation 2x^2-5=27 ?

The solutions to the equation 2x^2-5=27 are x=4,x=-4

Find the zeros of 2x^2-5=27

The zeros of 2x^2-5=27 are x=4,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 5 = 27

x = 4, x = - 4

Solution steps

2 x^{2} - 5 = 27

Move

5

to the right side

2 x^{2} = 32

Divide both sides by

2

x^{2} = 16

For

x^{2} = f (a)

the solutions are

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

x = 16, x = - 16

16 = 4

- 16 = - 4

x = 4, x = - 4

4 x^{2} - 16 x = 9

5 x^{2} - 8 x + 12 = 0

x^{2} + 3 x - 3 = 7 x + 9

x^{2} + 10 x - 12

x^{2} - 4 x = 77

What are the solutions to the equation 2x^2-5=27 ?
The solutions to the equation 2x^2-5=27 are x=4,x=-4
Find the zeros of 2x^2-5=27
The zeros of 2x^2-5=27 are x=4,x=-4