What are the solutions to the equation (x+1)(x-1)-3=0 ?

The solutions to the equation (x+1)(x-1)-3=0 are x=2,x=-2

Find the zeros of (x+1)(x-1)-3=0

The zeros of (x+1)(x-1)-3=0 are x=2,x=-2

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(x + 1) (x - 1) - 3 = 0

x = 2, x = - 2

Solution steps

(x + 1) (x - 1) - 3 = 0

Expand

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

x^{2} - 4 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 0 \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 4}{2 \cdot 1}, x_{2} = \frac{- 0 - 4}{2 \cdot 1}

x = \frac{- 0 + 4}{2 \cdot 1} : 2

x = \frac{- 0 - 4}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 2, x = - 2

3 x^{2} - 8 x + 13 = 0

6 m^{2} + 2 m - 4 = 0

x, x^{2} + p x + q = 0

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

x^{2} + 9 x - 10 = 0

What are the solutions to the equation (x+1)(x-1)-3=0 ?
The solutions to the equation (x+1)(x-1)-3=0 are x=2,x=-2
Find the zeros of (x+1)(x-1)-3=0
The zeros of (x+1)(x-1)-3=0 are x=2,x=-2