What are the solutions to the equation 6x^2=4+x ?

The solutions to the equation 6x^2=4+x are x=(1+sqrt(97))/(12),x=(1-sqrt(97))/(12)

Find the zeros of 6x^2=4+x

The zeros of 6x^2=4+x are x=(1+sqrt(97))/(12),x=(1-sqrt(97))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 x^{2} = 4 + x

x = \frac{1 + 97}{12}, x = \frac{1 - 97}{12}

Decimal Notation

x = 0.90407 \dots, x = - 0.73740 \dots

Solution steps

6 x^{2} = 4 + x

Move

x

to the left side

6 x^{2} - x = 4

Move

4

to the left side

6 x^{2} - x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 4 )}{2 \cdot 6}

(- 1)^{2} - 4 \cdot 6 (- 4) = 97

x_{1, 2} = \frac{- ( - 1 ) \pm 97}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 97}{2 \cdot 6}, x_{2} = \frac{- ( - 1 ) - 97}{2 \cdot 6}

x = \frac{- ( - 1 ) + 97}{2 \cdot 6} : \frac{1 + 97}{12}

x = \frac{- ( - 1 ) - 97}{2 \cdot 6} : \frac{1 - 97}{12}

The solutions to the quadratic equation are:

x = \frac{1 + 97}{12}, x = \frac{1 - 97}{12}

14 m^{2} n + 7 mn

\frac{1}{256} x^{4} - \frac{1}{625}

x > - 3 or x \leq 4

p^{2} + 11 p + 24 = 0

4 x^{2} + 11 x + 7

What are the solutions to the equation 6x^2=4+x ?
The solutions to the equation 6x^2=4+x are x=(1+sqrt(97))/(12),x=(1-sqrt(97))/(12)
Find the zeros of 6x^2=4+x
The zeros of 6x^2=4+x are x=(1+sqrt(97))/(12),x=(1-sqrt(97))/(12)