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

The solutions to the equation x^2-x/2 = 1/3-(2x)/3 are x= 1/2 ,x=-2/3

Find the zeros of x^2-x/2 = 1/3-(2x)/3

The zeros of x^2-x/2 = 1/3-(2x)/3 are x= 1/2 ,x=-2/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - \frac{x}{2} = \frac{1}{3} - \frac{2 x}{3}

x = \frac{1}{2}, x = - \frac{2}{3}

Decimal Notation

x = 0.5, x = - 0.66666 \dots

Solution steps

x^{2} - \frac{x}{2} = \frac{1}{3} - \frac{2 x}{3}

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

x^{2} \cdot 6 - \frac{x}{2} \cdot 6 = \frac{1}{3} \cdot 6 - \frac{2 x}{3} \cdot 6

Simplify

6 x^{2} - 3 x = 2 - 4 x

Move

4 x

to the left side

6 x^{2} + x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 6 (- 2) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 6}, x_{2} = \frac{- 1 - 7}{2 \cdot 6}

x = \frac{- 1 + 7}{2 \cdot 6} : \frac{1}{2}

x = \frac{- 1 - 7}{2 \cdot 6} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{2}{3}

0.5 x^{2} + x = 0

45 x + 2 = 180

- 6 (n - 8) = 4 (12 - 5 n) + 14 n

- 5 x - 5 < - 4 x - 4

20 - 10 t = 20

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