What are the solutions to the equation 4(x+1)^2-7=45 ?

The solutions to the equation 4(x+1)^2-7=45 are x=sqrt(13)-1,x=-sqrt(13)-1

Find the zeros of 4(x+1)^2-7=45

The zeros of 4(x+1)^2-7=45 are x=sqrt(13)-1,x=-sqrt(13)-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 (x + 1)^{2} - 7 = 45

x = 13 - 1, x = - 13 - 1

Decimal

x = 2.60555 \dots, x = - 4.60555 \dots

Solution steps

4 (x + 1)^{2} - 7 = 45

Move

7

to the right side

4 (x + 1)^{2} = 52

Divide both sides by

4

(x + 1)^{2} = 13

For

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

the solutions are

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

Solve

x + 1 = 13 : x = 13 - 1

Solve

x + 1 = - 13 : x = - 13 - 1

The solutions to the quadratic equation are:

x = 13 - 1, x = - 13 - 1

3 x^{2} + 231 = 54 x

0 = x^{2} - 18 x + 64

x^{2} + 6 x - 100 = 6 x

5 (x + 6)^{2} - 48 = - 28

- 3 x^{2} - 21 x - 30 = 0

What are the solutions to the equation 4(x+1)^2-7=45 ?
The solutions to the equation 4(x+1)^2-7=45 are x=sqrt(13)-1,x=-sqrt(13)-1
Find the zeros of 4(x+1)^2-7=45
The zeros of 4(x+1)^2-7=45 are x=sqrt(13)-1,x=-sqrt(13)-1