What are the solutions to the equation 2x-(x+1)2=3x^2-6 ?

The solutions to the equation 2x-(x+1)2=3x^2-6 are x= 2/(sqrt(3)),x=-2/(sqrt(3))

Find the zeros of 2x-(x+1)2=3x^2-6

The zeros of 2x-(x+1)2=3x^2-6 are x= 2/(sqrt(3)),x=-2/(sqrt(3))

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

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

Decimal Notation

x = 1.15470 \dots, x = - 1.15470 \dots

Solution steps

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

Expand

2 x - (x + 1) \cdot 2 : - 2

- 2 = 3 x^{2} - 6

Switch sides

3 x^{2} - 6 = - 2

Move

6

to the right side

3 x^{2} = 4

Divide both sides by

3

x^{2} = \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

\frac{4}{3} = \frac{2}{3}

- \frac{4}{3} = - \frac{2}{3}

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

7 x + 39 \geq 53

x + 49 = 9 x + 17

3 2^{- \frac{6}{5}}

2 (3 x - 4) + x - 1 = - 16

\frac{( \frac{π}{2} )}{3}

What are the solutions to the equation 2x-(x+1)2=3x^2-6 ?
The solutions to the equation 2x-(x+1)2=3x^2-6 are x= 2/(sqrt(3)),x=-2/(sqrt(3))
Find the zeros of 2x-(x+1)2=3x^2-6
The zeros of 2x-(x+1)2=3x^2-6 are x= 2/(sqrt(3)),x=-2/(sqrt(3))