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

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

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

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

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

x = 1, x = - 1

Solution steps

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

Expand

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

- 3 x^{2} + 3 = x^{2} - 1

Move

3

to the right side

- 3 x^{2} = x^{2} - 4

Move

x^{2}

to the left side

- 4 x^{2} = - 4

Divide both sides by

- 4

x^{2} = 1

For

x^{2} = f (a)

the solutions are

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

x = 1, x = - 1

1 = 1

- 1 = - 1

x = 1, x = - 1

- 27 = 9 (- 1 + x)

64 x^{3} - 343

3^{2} + 3^{2}

(m + 3) (5 m + 1) = 3 (m - 4) + 7

0.0000987 x^{2} - 0.009 x + 0.0613 = 0

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