What are the solutions to the equation (6x-3)(2x+4)=3 ?

The solutions to the equation (6x-3)(2x+4)=3 are x=(-3+sqrt(29))/4 ,x=-(3+sqrt(29))/4

Find the zeros of (6x-3)(2x+4)=3

The zeros of (6x-3)(2x+4)=3 are x=(-3+sqrt(29))/4 ,x=-(3+sqrt(29))/4

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(6 x - 3) (2 x + 4) = 3

x = \frac{- 3 + 29}{4}, x = - \frac{3 + 29}{4}

Decimal Notation

x = 0.59629 \dots, x = - 2.09629 \dots

Solution steps

(6 x - 3) (2 x + 4) = 3

Expand

(6 x - 3) (2 x + 4) : 12 x^{2} + 18 x - 12

12 x^{2} + 18 x - 12 = 3

Move

3

to the left side

12 x^{2} + 18 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 12 ( - 15 )}{2 \cdot 12}

1 8^{2} - 4 \cdot 12 (- 15) = 629

x_{1, 2} = \frac{- 18 \pm 6 29}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 18 + 6 29}{2 \cdot 12}, x_{2} = \frac{- 18 - 6 29}{2 \cdot 12}

x = \frac{- 18 + 6 29}{2 \cdot 12} : \frac{- 3 + 29}{4}

x = \frac{- 18 - 6 29}{2 \cdot 12} : - \frac{3 + 29}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 29}{4}, x = - \frac{3 + 29}{4}

4 x^{2} - 2 x + 10 = 0

x^{2} + 9 x = 27

2 x^{2} - 10 x + 9 = 0

100 x^{2} = 1

3 x^{2} + 9 x - 26 = 4 x^{2} - x - 1

What are the solutions to the equation (6x-3)(2x+4)=3 ?
The solutions to the equation (6x-3)(2x+4)=3 are x=(-3+sqrt(29))/4 ,x=-(3+sqrt(29))/4
Find the zeros of (6x-3)(2x+4)=3
The zeros of (6x-3)(2x+4)=3 are x=(-3+sqrt(29))/4 ,x=-(3+sqrt(29))/4