What are the solutions to the equation 2(2x-4)^2-41=29 ?

The solutions to the equation 2(2x-4)^2-41=29 are x=(sqrt(35)+4)/2 ,x=(-sqrt(35)+4)/2

Find the zeros of 2(2x-4)^2-41=29

The zeros of 2(2x-4)^2-41=29 are x=(sqrt(35)+4)/2 ,x=(-sqrt(35)+4)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 (2 x - 4)^{2} - 41 = 29

x = \frac{35 + 4}{2}, x = \frac{- 35 + 4}{2}

Decimal Notation

x = 4.95803 \dots, x = - 0.95803 \dots

Solution steps

2 (2 x - 4)^{2} - 41 = 29

Move

41

to the right side

2 (2 x - 4)^{2} = 70

Divide both sides by

2

(2 x - 4)^{2} = 35

For

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

the solutions are

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

Solve

2 x - 4 = 35 : x = \frac{35 + 4}{2}

Solve

2 x - 4 = - 35 : x = \frac{- 35 + 4}{2}

The solutions to the quadratic equation are:

x = \frac{35 + 4}{2}, x = \frac{- 35 + 4}{2}

r^{2} - 6 r + 18 = 0

x^{2} + 26 = - 12 x

0 = 16.1 x^{2} + 70 x - 580

2 x^{2} + 12 x + 36 = 0

- 6 x^{2} + 60 x - 144 = 0

What are the solutions to the equation 2(2x-4)^2-41=29 ?
The solutions to the equation 2(2x-4)^2-41=29 are x=(sqrt(35)+4)/2 ,x=(-sqrt(35)+4)/2
Find the zeros of 2(2x-4)^2-41=29
The zeros of 2(2x-4)^2-41=29 are x=(sqrt(35)+4)/2 ,x=(-sqrt(35)+4)/2