What are the solutions to the equation (x+3)(x-3)=55 ?

The solutions to the equation (x+3)(x-3)=55 are x=8,x=-8

Find the zeros of (x+3)(x-3)=55

The zeros of (x+3)(x-3)=55 are x=8,x=-8

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(x + 3) (x - 3) = 55

x = 8, x = - 8

Solution steps

(x + 3) (x - 3) = 55

Expand

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

x^{2} - 9 = 55

Move

55

to the left side

x^{2} - 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 64 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 64) = 16

x_{1, 2} = \frac{- 0 \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 16}{2 \cdot 1}, x_{2} = \frac{- 0 - 16}{2 \cdot 1}

x = \frac{- 0 + 16}{2 \cdot 1} : 8

x = \frac{- 0 - 16}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 8, x = - 8

(\frac{5}{13})^{2} - (\frac{12}{13})^{2}

lo g_{10} (25) + lo g_{10} (4)

\frac{2 - 1}{2 + 1} + \frac{4}{2}

\frac{x ^{4}}{x ^{4}}

\frac{- 2}{- 10}

What are the solutions to the equation (x+3)(x-3)=55 ?
The solutions to the equation (x+3)(x-3)=55 are x=8,x=-8
Find the zeros of (x+3)(x-3)=55
The zeros of (x+3)(x-3)=55 are x=8,x=-8