What are the solutions to the equation 3=(W-7)(W-9) ?

The solutions to the equation 3=(W-7)(W-9) are W=10,W=6

Find the zeros of 3=(W-7)(W-9)

The zeros of 3=(W-7)(W-9) are W=10,W=6

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3 = (W - 7) (W - 9)

W = 10, W = 6

Solution steps

3 = (W - 7) (W - 9)

Expand

(W - 7) (W - 9) : W^{2} - 16 W + 63

3 = W^{2} - 16 W + 63

Switch sides

W^{2} - 16 W + 63 = 3

Move

3

to the left side

W^{2} - 16 W + 60 = 0

Solve with the quadratic formula

W_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 60}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 60 = 4

W_{1, 2} = \frac{- ( - 16 ) \pm 4}{2 \cdot 1}

Separate the solutions

W_{1} = \frac{- ( - 16 ) + 4}{2 \cdot 1}, W_{2} = \frac{- ( - 16 ) - 4}{2 \cdot 1}

W = \frac{- ( - 16 ) + 4}{2 \cdot 1} : 10

W = \frac{- ( - 16 ) - 4}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

W = 10, W = 6

\frac{3 + x}{4} + \frac{2 - x}{3} < 0

x^{2} y^{2} - 7 x y + 10

7 v - 15 = 3 (8 - 2 v)

(\frac{2}{3})^{2} \cdot (\frac{2}{3})^{3}

81 - z^{2}

What are the solutions to the equation 3=(W-7)(W-9) ?
The solutions to the equation 3=(W-7)(W-9) are W=10,W=6
Find the zeros of 3=(W-7)(W-9)
The zeros of 3=(W-7)(W-9) are W=10,W=6