What are the solutions to the equation (w-8)(w+5)=14 ?

The solutions to the equation (w-8)(w+5)=14 are w=9,w=-6

Find the zeros of (w-8)(w+5)=14

The zeros of (w-8)(w+5)=14 are w=9,w=-6

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(w - 8) (w + 5) = 14

w = 9, w = - 6

Solution steps

(w - 8) (w + 5) = 14

Expand

(w - 8) (w + 5) : w^{2} - 3 w - 40

w^{2} - 3 w - 40 = 14

Move

14

to the left side

w^{2} - 3 w - 54 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 54 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 54) = 15

w_{1, 2} = \frac{- ( - 3 ) \pm 15}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 3 ) + 15}{2 \cdot 1}, w_{2} = \frac{- ( - 3 ) - 15}{2 \cdot 1}

w = \frac{- ( - 3 ) + 15}{2 \cdot 1} : 9

w = \frac{- ( - 3 ) - 15}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

w = 9, w = - 6

6 x^{2} - 8 x + 1

(r - 7)^{2} = 81

2 x^{2} - 7 x + 9 = (x - 3) (x + 1) + 3 x

2 x (x + 3) + 1 = 0

(x + 2)^{3} = x (x^{2} - 10 x + 25)

What are the solutions to the equation (w-8)(w+5)=14 ?
The solutions to the equation (w-8)(w+5)=14 are w=9,w=-6
Find the zeros of (w-8)(w+5)=14
The zeros of (w-8)(w+5)=14 are w=9,w=-6