What are the solutions to the equation (3x-1)(2x+3)=8 ?

The solutions to the equation (3x-1)(2x+3)=8 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)

Find the zeros of (3x-1)(2x+3)=8

The zeros of (3x-1)(2x+3)=8 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)

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(3 x - 1) (2 x + 3) = 8

x = \frac{- 7 + 313}{12}, x = \frac{- 7 - 313}{12}

Decimal Notation

x = 0.89098 \dots, x = - 2.05765 \dots

Solution steps

(3 x - 1) (2 x + 3) = 8

Expand

(3 x - 1) (2 x + 3) : 6 x^{2} + 7 x - 3

6 x^{2} + 7 x - 3 = 8

Move

8

to the left side

6 x^{2} + 7 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 ( - 11 )}{2 \cdot 6}

7^{2} - 4 \cdot 6 (- 11) = 313

x_{1, 2} = \frac{- 7 \pm 313}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 7 + 313}{2 \cdot 6}, x_{2} = \frac{- 7 - 313}{2 \cdot 6}

x = \frac{- 7 + 313}{2 \cdot 6} : \frac{- 7 + 313}{12}

x = \frac{- 7 - 313}{2 \cdot 6} : \frac{- 7 - 313}{12}

The solutions to the quadratic equation are:

x = \frac{- 7 + 313}{12}, x = \frac{- 7 - 313}{12}

2 m^{2} + m = 1

1.5 t^{2} - 6 t + 3 = 0

t^{2} - t - 6 = 4 (t - 3)

6 x^{2} - 9 x - 15 = 0

2 x (3 x - 5) = x

What are the solutions to the equation (3x-1)(2x+3)=8 ?
The solutions to the equation (3x-1)(2x+3)=8 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)
Find the zeros of (3x-1)(2x+3)=8
The zeros of (3x-1)(2x+3)=8 are x=(-7+sqrt(313))/(12),x=(-7-sqrt(313))/(12)