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

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

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

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

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

x = 6, x = - 4

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} : 6

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

The solutions to the quadratic equation are:

x = 6, x = - 4

9 x^{2} + 21 x - 8 = 0

(3 x - 2) x = (5 x - 1) x

2 (x + 1)^{2} + 30 = 32

33 b = 10 - 7 b^{2}

12 x^{2} - 288 x + 1260 = 0

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