What are the solutions to the equation (2x+1)^2-128=0 ?

The solutions to the equation (2x+1)^2-128=0 are x=(8sqrt(2)-1)/2 ,x=(-8sqrt(2)-1)/2

Find the zeros of (2x+1)^2-128=0

The zeros of (2x+1)^2-128=0 are x=(8sqrt(2)-1)/2 ,x=(-8sqrt(2)-1)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x + 1)^{2} - 128 = 0

x = \frac{8 2 - 1}{2}, x = \frac{- 8 2 - 1}{2}

Solution steps

(2 x + 1)^{2} - 128 = 0

Move

128

to the right side

(2 x + 1)^{2} = 128

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

2 x + 1 = 128 : x = \frac{8 2 - 1}{2}

Solve

2 x + 1 = - 128 : x = \frac{- 8 2 - 1}{2}

The solutions to the quadratic equation are:

x = \frac{8 2 - 1}{2}, x = \frac{- 8 2 - 1}{2}

2 (a - 4 x) + a x (3 x 2 + 4) = 2 (6 x 3 + 5)

6 x^{2} - 12 x + 390 = 0

x, x^{2} - 12 x - 36 = 0

5 x + 2 = - 4 x^{2}

0.225375 = 0.000015 (x + 100)^{2}

What are the solutions to the equation (2x+1)^2-128=0 ?
The solutions to the equation (2x+1)^2-128=0 are x=(8sqrt(2)-1)/2 ,x=(-8sqrt(2)-1)/2
Find the zeros of (2x+1)^2-128=0
The zeros of (2x+1)^2-128=0 are x=(8sqrt(2)-1)/2 ,x=(-8sqrt(2)-1)/2