What are the solutions to the equation (3x+1)(x-1)=15 ?

The solutions to the equation (3x+1)(x-1)=15 are x= 8/3 ,x=-2

Find the zeros of (3x+1)(x-1)=15

The zeros of (3x+1)(x-1)=15 are x= 8/3 ,x=-2

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(3 x + 1) (x - 1) = 15

x = \frac{8}{3}, x = - 2

Decimal Notation

x = 2.66666 \dots, x = - 2

Solution steps

(3 x + 1) (x - 1) = 15

Expand

(3 x + 1) (x - 1) : 3 x^{2} - 2 x - 1

3 x^{2} - 2 x - 1 = 15

Move

15

to the left side

3 x^{2} - 2 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 16 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 16) = 14

x_{1, 2} = \frac{- ( - 2 ) \pm 14}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 14}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 14}{2 \cdot 3}

x = \frac{- ( - 2 ) + 14}{2 \cdot 3} : \frac{8}{3}

x = \frac{- ( - 2 ) - 14}{2 \cdot 3} : - 2

The solutions to the quadratic equation are:

x = \frac{8}{3}, x = - 2

b^{8} = x^{2} + 4 x + b

(\frac{1}{2} - x) (- \frac{1}{2} - x) = 0

x, x^{2} = e^{y}

x, x^{2} - 9 x - 22 = 0

- 2 s^{2} - 60 s - 310 = 0

What are the solutions to the equation (3x+1)(x-1)=15 ?
The solutions to the equation (3x+1)(x-1)=15 are x= 8/3 ,x=-2
Find the zeros of (3x+1)(x-1)=15
The zeros of (3x+1)(x-1)=15 are x= 8/3 ,x=-2