What are the solutions to the equation (x-3)(x-5)=24 ?

The solutions to the equation (x-3)(x-5)=24 are x=9,x=-1

Find the zeros of (x-3)(x-5)=24

The zeros of (x-3)(x-5)=24 are x=9,x=-1

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(x - 3) (x - 5) = 24

x = 9, x = - 1

Solution steps

(x - 3) (x - 5) = 24

Expand

(x - 3) (x - 5) : x^{2} - 8 x + 15

x^{2} - 8 x + 15 = 24

Move

24

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 9) = 10

x_{1, 2} = \frac{- ( - 8 ) \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 10}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 10}{2 \cdot 1}

x = \frac{- ( - 8 ) + 10}{2 \cdot 1} : 9

x = \frac{- ( - 8 ) - 10}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 9, x = - 1

x^{2} - 4.1 x + 2.2 = 0

48 x^{2} + 22 x - 5 = 0

X^{2} - X = 0

(2 m + 5) x^{2} + 3 m x - 1 = - x^{2} + 8

x^{(2)} + 12 x + 36 = 0

What are the solutions to the equation (x-3)(x-5)=24 ?
The solutions to the equation (x-3)(x-5)=24 are x=9,x=-1
Find the zeros of (x-3)(x-5)=24
The zeros of (x-3)(x-5)=24 are x=9,x=-1