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

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

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

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

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

b = \frac{2}{3}, b = - \frac{2}{5}

Decimal Notation

b = 0.66666 \dots, b = - 0.4

Solution steps

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

Expand

(5 b - 3) (3 b + 1) : 15 b^{2} - 4 b - 3

15 b^{2} - 4 b - 3 = 1

Move

1

to the left side

15 b^{2} - 4 b - 4 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 15 ( - 4 )}{2 \cdot 15}

(- 4)^{2} - 4 \cdot 15 (- 4) = 16

b_{1, 2} = \frac{- ( - 4 ) \pm 16}{2 \cdot 15}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 16}{2 \cdot 15}, b_{2} = \frac{- ( - 4 ) - 16}{2 \cdot 15}

b = \frac{- ( - 4 ) + 16}{2 \cdot 15} : \frac{2}{3}

b = \frac{- ( - 4 ) - 16}{2 \cdot 15} : - \frac{2}{5}

The solutions to the quadratic equation are:

b = \frac{2}{3}, b = - \frac{2}{5}

4 a^{2} + 8 a b^{2} + 4 b^{4}

a^{2} + 9 a + 20

\frac{3}{5} + \frac{1}{4} z + \frac{1}{4} < z

3 (- 4)^{2}

0 = 2 x^{2} + 4 x

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