What are the solutions to the equation (x+3)(x-3)=7 ?

The solutions to the equation (x+3)(x-3)=7 are x=4,x=-4

Find the zeros of (x+3)(x-3)=7

The zeros of (x+3)(x-3)=7 are x=4,x=-4

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(x + 3) (x - 3) = 7

x = 4, x = - 4

Solution steps

(x + 3) (x - 3) = 7

Expand

(x + 3) (x - 3) : x^{2} - 9

x^{2} - 9 = 7

Move

7

to the left side

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

9 (x - 2) = 18

12 x + 4 = 20 x - 12

- \frac{19}{2} (\frac{1}{100}) + \frac{19}{2}

n - 2 = \frac{n ( n - 3 )}{2}

\frac{π 6 \cdot 4 5 ^{\circ}}{18 0 ^{\circ}}

What are the solutions to the equation (x+3)(x-3)=7 ?
The solutions to the equation (x+3)(x-3)=7 are x=4,x=-4
Find the zeros of (x+3)(x-3)=7
The zeros of (x+3)(x-3)=7 are x=4,x=-4