What are the solutions to the equation (x+2)(x-2)=32 ?

The solutions to the equation (x+2)(x-2)=32 are x=6,x=-6

Find the zeros of (x+2)(x-2)=32

The zeros of (x+2)(x-2)=32 are x=6,x=-6

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(x + 2) (x - 2) = 32

x = 6, x = - 6

Solution steps

(x + 2) (x - 2) = 32

Expand

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

x^{2} - 4 = 32

Move

32

to the left side

x^{2} - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 36) = 12

x_{1, 2} = \frac{- 0 \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 12}{2 \cdot 1}, x_{2} = \frac{- 0 - 12}{2 \cdot 1}

x = \frac{- 0 + 12}{2 \cdot 1} : 6

x = \frac{- 0 - 12}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 6, x = - 6

(x + 2)^{2} - 2 (x + 2) - 15 = 0

x^{2} + x - 12

9 x^{2} + 18 x - 7 = 0

x^{2} + 4 x - 60 = 0

50 y^{2} - 350 y + 600 = 0

What are the solutions to the equation (x+2)(x-2)=32 ?
The solutions to the equation (x+2)(x-2)=32 are x=6,x=-6
Find the zeros of (x+2)(x-2)=32
The zeros of (x+2)(x-2)=32 are x=6,x=-6