What are the solutions to the equation (x-3)(x+5)=9 ?

The solutions to the equation (x-3)(x+5)=9 are x=4,x=-6

Find the zeros of (x-3)(x+5)=9

The zeros of (x-3)(x+5)=9 are x=4,x=-6

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(x - 3) (x + 5) = 9

x = 4, x = - 6

Solution steps

(x - 3) (x + 5) = 9

Expand

(x - 3) (x + 5) : x^{2} + 2 x - 15

x^{2} + 2 x - 15 = 9

Move

9

to the left side

x^{2} + 2 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 1}, x_{2} = \frac{- 2 - 10}{2 \cdot 1}

x = \frac{- 2 + 10}{2 \cdot 1} : 4

x = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

\frac{x - 7}{4} = - 2

\frac{1 - 4 x}{3} - \frac{1 + 3 x}{8} = \frac{x}{6}

\frac{1}{2} + 10

\frac{15 x ^{2} y ^{5}}{25 x ^{3} y}

(1 - x)^{- 1}

What are the solutions to the equation (x-3)(x+5)=9 ?
The solutions to the equation (x-3)(x+5)=9 are x=4,x=-6
Find the zeros of (x-3)(x+5)=9
The zeros of (x-3)(x+5)=9 are x=4,x=-6