What are the solutions to the equation 2(x+7)^2+27=41 ?

The solutions to the equation 2(x+7)^2+27=41 are x=sqrt(7)-7,x=-sqrt(7)-7

Find the zeros of 2(x+7)^2+27=41

The zeros of 2(x+7)^2+27=41 are x=sqrt(7)-7,x=-sqrt(7)-7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 (x + 7)^{2} + 27 = 41

x = 7 - 7, x = - 7 - 7

Decimal

x = - 4.35424 \dots, x = - 9.64575 \dots

Solution steps

2 (x + 7)^{2} + 27 = 41

Move

27

to the right side

2 (x + 7)^{2} = 14

Divide both sides by

2

(x + 7)^{2} = 7

For

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

the solutions are

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

Solve

x + 7 = 7 : x = 7 - 7

Solve

x + 7 = - 7 : x = - 7 - 7

The solutions to the quadratic equation are:

x = 7 - 7, x = - 7 - 7

2 (x^{2} - 9) + 1 = 1

3 x (x - 3) = 2 (x + 2)

2 x^{2} - 200 = 138

x^{2} - 8 x + 15 = 0

x^{2} = 64

What are the solutions to the equation 2(x+7)^2+27=41 ?
The solutions to the equation 2(x+7)^2+27=41 are x=sqrt(7)-7,x=-sqrt(7)-7
Find the zeros of 2(x+7)^2+27=41
The zeros of 2(x+7)^2+27=41 are x=sqrt(7)-7,x=-sqrt(7)-7