What are the solutions to the equation (9x-1)(2x+1)=2x ?

The solutions to the equation (9x-1)(2x+1)=2x are x=(-5+sqrt(97))/(36),x=(-5-sqrt(97))/(36)

Find the zeros of (9x-1)(2x+1)=2x

The zeros of (9x-1)(2x+1)=2x are x=(-5+sqrt(97))/(36),x=(-5-sqrt(97))/(36)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(9 x - 1) (2 x + 1) = 2 x

x = \frac{- 5 + 97}{36}, x = \frac{- 5 - 97}{36}

Decimal Notation

x = 0.13469 \dots, x = - 0.41246 \dots

Solution steps

(9 x - 1) (2 x + 1) = 2 x

Expand

(9 x - 1) (2 x + 1) : 18 x^{2} + 7 x - 1

18 x^{2} + 7 x - 1 = 2 x

Move

2 x

to the left side

18 x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 18 ( - 1 )}{2 \cdot 18}

5^{2} - 4 \cdot 18 (- 1) = 97

x_{1, 2} = \frac{- 5 \pm 97}{2 \cdot 18}

Separate the solutions

x_{1} = \frac{- 5 + 97}{2 \cdot 18}, x_{2} = \frac{- 5 - 97}{2 \cdot 18}

x = \frac{- 5 + 97}{2 \cdot 18} : \frac{- 5 + 97}{36}

x = \frac{- 5 - 97}{2 \cdot 18} : \frac{- 5 - 97}{36}

The solutions to the quadratic equation are:

x = \frac{- 5 + 97}{36}, x = \frac{- 5 - 97}{36}

2 x^{2} - 3 x = 405

3 v^{2} + 11 v + 9 = 0

2 = x (x + 2)

2 x (x - 2) = 5 x^{2} - 5 x

(3 a + 12) (3 a - 12) = 0

What are the solutions to the equation (9x-1)(2x+1)=2x ?
The solutions to the equation (9x-1)(2x+1)=2x are x=(-5+sqrt(97))/(36),x=(-5-sqrt(97))/(36)
Find the zeros of (9x-1)(2x+1)=2x
The zeros of (9x-1)(2x+1)=2x are x=(-5+sqrt(97))/(36),x=(-5-sqrt(97))/(36)