What are the solutions to the equation (2x+3)(3x-1)=4 ?

The solutions to the equation (2x+3)(3x-1)=4 are x=(-7+sqrt(217))/(12),x=(-7-sqrt(217))/(12)

Find the zeros of (2x+3)(3x-1)=4

The zeros of (2x+3)(3x-1)=4 are x=(-7+sqrt(217))/(12),x=(-7-sqrt(217))/(12)

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(2 x + 3) (3 x - 1) = 4

x = \frac{- 7 + 217}{12}, x = \frac{- 7 - 217}{12}

Decimal Notation

x = 0.64424 \dots, x = - 1.81090 \dots

Solution steps

(2 x + 3) (3 x - 1) = 4

Expand

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

6 x^{2} + 7 x - 3 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 6 (- 7) = 217

x_{1, 2} = \frac{- 7 \pm 217}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 7 + 217}{2 \cdot 6}, x_{2} = \frac{- 7 - 217}{2 \cdot 6}

x = \frac{- 7 + 217}{2 \cdot 6} : \frac{- 7 + 217}{12}

x = \frac{- 7 - 217}{2 \cdot 6} : \frac{- 7 - 217}{12}

The solutions to the quadratic equation are:

x = \frac{- 7 + 217}{12}, x = \frac{- 7 - 217}{12}

- 16 t^{2} + 152 t + 8 = 0

3 + 7 + 11 + (4 n - 1) = n (2 n + 1)

7 - x^{2} = 0

6 x^{4} + 4 x^{2} - 3 x + 2 = 2 x^{2} (3 x^{2} - 1)

6 k^{2} - 5 k - 6 = 0

What are the solutions to the equation (2x+3)(3x-1)=4 ?
The solutions to the equation (2x+3)(3x-1)=4 are x=(-7+sqrt(217))/(12),x=(-7-sqrt(217))/(12)
Find the zeros of (2x+3)(3x-1)=4
The zeros of (2x+3)(3x-1)=4 are x=(-7+sqrt(217))/(12),x=(-7-sqrt(217))/(12)