What are the solutions to the equation 4(x-4)^2-33=-5 ?

The solutions to the equation 4(x-4)^2-33=-5 are x=sqrt(7)+4,x=-sqrt(7)+4

Find the zeros of 4(x-4)^2-33=-5

The zeros of 4(x-4)^2-33=-5 are x=sqrt(7)+4,x=-sqrt(7)+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 (x - 4)^{2} - 33 = - 5

x = 7 + 4, x = - 7 + 4

Decimal Notation

x = 6.64575 \dots, x = 1.35424 \dots

Solution steps

4 (x - 4)^{2} - 33 = - 5

Move

33

to the right side

4 (x - 4)^{2} = 28

Divide both sides by

4

(x - 4)^{2} = 7

For

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

the solutions are

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

Solve

x - 4 = 7 : x = 7 + 4

Solve

x - 4 = - 7 : x = - 7 + 4

The solutions to the quadratic equation are:

x = 7 + 4, x = - 7 + 4

2 v^{2} + 15 v + 19 = 0

0 = 2 x^{2} - 10 x + 7

- 18 x^{2} + 45 x - 25 = 0

a^{2} - 8 a + 12 = 2

- 3 (x + 2)^{2} = 24

What are the solutions to the equation 4(x-4)^2-33=-5 ?
The solutions to the equation 4(x-4)^2-33=-5 are x=sqrt(7)+4,x=-sqrt(7)+4
Find the zeros of 4(x-4)^2-33=-5
The zeros of 4(x-4)^2-33=-5 are x=sqrt(7)+4,x=-sqrt(7)+4