What are the solutions to the equation (x-2)(x+2)-12=0 ?

The solutions to the equation (x-2)(x+2)-12=0 are x=4,x=-4

Find the zeros of (x-2)(x+2)-12=0

The zeros of (x-2)(x+2)-12=0 are x=4,x=-4

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(x - 2) (x + 2) - 12 = 0

x = 4, x = - 4

Solution steps

(x - 2) (x + 2) - 12 = 0

Expand

(x - 2) (x + 2) - 12 : x^{2} - 16

x^{2} - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 16) = 8

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

Separate the solutions

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

x = \frac{- 0 + 8}{2 \cdot 1} : 4

x = \frac{- 0 - 8}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 4, x = - 4

x^{2} + 29 = 0

a^{2} + 4^{2} = 7^{2}

4 - x^{2} - 4 + 4 x = 0

a^{2} + 8 a + 7 = 0

x^{2} - 16

What are the solutions to the equation (x-2)(x+2)-12=0 ?
The solutions to the equation (x-2)(x+2)-12=0 are x=4,x=-4
Find the zeros of (x-2)(x+2)-12=0
The zeros of (x-2)(x+2)-12=0 are x=4,x=-4