What are the solutions to the equation (w-7)(w+2)=-18 ?

The solutions to the equation (w-7)(w+2)=-18 are w=4,w=1

Find the zeros of (w-7)(w+2)=-18

The zeros of (w-7)(w+2)=-18 are w=4,w=1

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(w - 7) (w + 2) = - 18

w = 4, w = 1

Solution steps

(w - 7) (w + 2) = - 18

Expand

(w - 7) (w + 2) : w^{2} - 5 w - 14

w^{2} - 5 w - 14 = - 18

Move

18

to the left side

w^{2} - 5 w + 4 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

w_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, w_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

w = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

w = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

w = 4, w = 1

2 y^{2} + 7 y = - 3

16 x^{2} + 49 = 0

3 x^{2} - 21 = 2 x

m^{2} + 100 = 0

x^{2} + 11 x + 13 = 0

What are the solutions to the equation (w-7)(w+2)=-18 ?
The solutions to the equation (w-7)(w+2)=-18 are w=4,w=1
Find the zeros of (w-7)(w+2)=-18
The zeros of (w-7)(w+2)=-18 are w=4,w=1