What are the solutions to the equation (x+3)(x-3)=27 ?

The solutions to the equation (x+3)(x-3)=27 are x=6,x=-6

Find the zeros of (x+3)(x-3)=27

The zeros of (x+3)(x-3)=27 are x=6,x=-6

Popular >

(x + 3) (x - 3) = 27

x = 6, x = - 6

Solution steps

(x + 3) (x - 3) = 27

Expand

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

x^{2} - 9 = 27

Move

27

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

30 + 40 x - 3 x^{2} = 0

\frac{1}{d ^{2}} = \frac{h ^{2} + k ^{2}}{a ^{2}} + \frac{l ^{2}}{c ^{2}}

8 x^{2} + 16 x + 3 = 0

x, x^{2} + y^{2} - 4 x - 2 y + 85 = 0

d, d^{2} y - 4 e^{5 x} = 0

What are the solutions to the equation (x+3)(x-3)=27 ?
The solutions to the equation (x+3)(x-3)=27 are x=6,x=-6
Find the zeros of (x+3)(x-3)=27
The zeros of (x+3)(x-3)=27 are x=6,x=-6