What are the solutions to the equation (x/8)^2+12=x ?

The solutions to the equation (x/8)^2+12=x are x=48,x=16

Find the zeros of (x/8)^2+12=x

The zeros of (x/8)^2+12=x are x=48,x=16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\frac{x}{8})^{2} + 12 = x

x = 48, x = 16

Solution steps

(\frac{x}{8})^{2} + 12 = x

Expand

(\frac{x}{8})^{2} + 12 : \frac{x ^{2}}{64} + 12

\frac{x ^{2}}{64} + 12 = x

Move

x

to the left side

\frac{x ^{2}}{64} + 12 - x = 0

Write in the standard form

a x^{2} + b x + c = 0

\frac{1 \cdot x ^{2}}{64} - x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot \frac{1}{64} \cdot 12}{2 \cdot \frac{1}{64}}

(- 1)^{2} - 4 \cdot \frac{1}{64} \cdot 12 = \frac{1}{2}

x_{1, 2} = \frac{- ( - 1 ) \pm \frac{1}{2}}{2 \cdot \frac{1}{64}}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + \frac{1}{2}}{2 \cdot \frac{1}{64}}, x_{2} = \frac{- ( - 1 ) - \frac{1}{2}}{2 \cdot \frac{1}{64}}

x = \frac{- ( - 1 ) + \frac{1}{2}}{2 \frac{1}{64}} : 48

x = \frac{- ( - 1 ) - \frac{1}{2}}{2 \frac{1}{64}} : 16

The solutions to the quadratic equation are:

x = 48, x = 16

x^{2} + 7 = 5 x + 1

x^{2} - \frac{11}{35} x - \frac{2}{7} = 0

x^{2} - x + 8 = - 10 x

x, x^{2} + y^{2} + 6 x - 2 y + 16 = 0

2 x^{2} + x - 3 = - 2 x^{2} + x + 1

What are the solutions to the equation (x/8)^2+12=x ?
The solutions to the equation (x/8)^2+12=x are x=48,x=16
Find the zeros of (x/8)^2+12=x
The zeros of (x/8)^2+12=x are x=48,x=16