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

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

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

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

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

x = \frac{1}{3}, x = - \frac{1}{3}

Decimal Notation

x = 0.57735 \dots, x = - 0.57735 \dots

Solution steps

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

Expand

2 + 3 (2 x + 1) (2 x - 1) : 12 x^{2} - 1

12 x^{2} - 1 = 3

Move

3

to the left side

12 x^{2} - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 12 ( - 4 )}{2 \cdot 12}

0^{2} - 4 \cdot 12 (- 4) = 83

x_{1, 2} = \frac{- 0 \pm 8 3}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 0 + 8 3}{2 \cdot 12}, x_{2} = \frac{- 0 - 8 3}{2 \cdot 12}

x = \frac{- 0 + 8 3}{2 \cdot 12} : \frac{1}{3}

x = \frac{- 0 - 8 3}{2 \cdot 12} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = - \frac{1}{3}

10 - 4 x - x^{2}

x^{2} - 4 x - 20 = - 5 x

(2 - x)^{2} - (2 x - 3)^{2} = 50

(5 - x)^{2} - (3 x - 1)^{2} = 0

(y + 2)^{2} = - 9

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