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

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

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

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

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

x = - \frac{1}{2 3}, x = \frac{1}{2 3}

Decimal Notation

x = - 0.28867 \dots, x = 0.28867 \dots

Solution steps

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

Expand

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

- 12 x^{2} + 3 = 2

Move

2

to the left side

- 12 x^{2} + 1 = 0

Solve with the quadratic formula

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

0^{2} - 4 (- 12) \cdot 1 = 43

x_{1, 2} = \frac{- 0 \pm 4 3}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- 0 + 4 3}{2 ( - 12 )}, x_{2} = \frac{- 0 - 4 3}{2 ( - 12 )}

x = \frac{- 0 + 4 3}{2 ( - 12 )} : - \frac{1}{2 3}

x = \frac{- 0 - 4 3}{2 ( - 12 )} : \frac{1}{2 3}

The solutions to the quadratic equation are:

x = - \frac{1}{2 3}, x = \frac{1}{2 3}

6 - 7 x^{2} = 7

x^{2} - x = \frac{( x + 1 ) ( x + 3 )}{2} - 2

n^{2} + 5 n + 7 = 7

x, 5 x^{2} + 2 y^{2} = 0

x, f (a + b) = x^{2} - 4 x

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