What are the solutions to the equation (3b-1)(2b+5)=3 ?

The solutions to the equation (3b-1)(2b+5)=3 are b= 1/2 ,b=-8/3

Find the zeros of (3b-1)(2b+5)=3

The zeros of (3b-1)(2b+5)=3 are b= 1/2 ,b=-8/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3 b - 1) (2 b + 5) = 3

b = \frac{1}{2}, b = - \frac{8}{3}

Decimal Notation

b = 0.5, b = - 2.66666 \dots

Solution steps

(3 b - 1) (2 b + 5) = 3

Expand

(3 b - 1) (2 b + 5) : 6 b^{2} + 13 b - 5

6 b^{2} + 13 b - 5 = 3

Move

3

to the left side

6 b^{2} + 13 b - 8 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 6 ( - 8 )}{2 \cdot 6}

1 3^{2} - 4 \cdot 6 (- 8) = 19

b_{1, 2} = \frac{- 13 \pm 19}{2 \cdot 6}

Separate the solutions

b_{1} = \frac{- 13 + 19}{2 \cdot 6}, b_{2} = \frac{- 13 - 19}{2 \cdot 6}

b = \frac{- 13 + 19}{2 \cdot 6} : \frac{1}{2}

b = \frac{- 13 - 19}{2 \cdot 6} : - \frac{8}{3}

The solutions to the quadratic equation are:

b = \frac{1}{2}, b = - \frac{8}{3}

5 x^{2} = 16 x + 16

- 3 x^{2} + 4 x = - 8

20 = y^{2}

40 = (x + 20) (x - 20)

0 = - 3 x^{2}

What are the solutions to the equation (3b-1)(2b+5)=3 ?
The solutions to the equation (3b-1)(2b+5)=3 are b= 1/2 ,b=-8/3
Find the zeros of (3b-1)(2b+5)=3
The zeros of (3b-1)(2b+5)=3 are b= 1/2 ,b=-8/3