What are the solutions to the equation 3(x-4)^2+1=13 ?

The solutions to the equation 3(x-4)^2+1=13 are x=6,x=2

Find the zeros of 3(x-4)^2+1=13

The zeros of 3(x-4)^2+1=13 are x=6,x=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 (x - 4)^{2} + 1 = 13

x = 6, x = 2

Solution steps

3 (x - 4)^{2} + 1 = 13

Move

1

to the right side

3 (x - 4)^{2} = 12

Divide both sides by

3

(x - 4)^{2} = 4

For

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

the solutions are

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

Solve

x - 4 = 4 : x = 6

Solve

x - 4 = - 4 : x = 2

The solutions to the quadratic equation are:

x = 6, x = 2

\frac{5 x}{3} + \frac{3 x}{5} = \frac{4}{15}

(\frac{2 r ^{3} t ^{6}}{5 u ^{9}})^{4}

\frac{25 !}{5 ! 20 !}

16 - 2 t = 23 t + 9

r^{2} + 12 r = - 32

What are the solutions to the equation 3(x-4)^2+1=13 ?
The solutions to the equation 3(x-4)^2+1=13 are x=6,x=2
Find the zeros of 3(x-4)^2+1=13
The zeros of 3(x-4)^2+1=13 are x=6,x=2