What are the solutions to the equation (w-3)^2=29-5w ?

The solutions to the equation (w-3)^2=29-5w are w=5,w=-4

Find the zeros of (w-3)^2=29-5w

The zeros of (w-3)^2=29-5w are w=5,w=-4

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(w - 3)^{2} = 29 - 5 w

w = 5, w = - 4

Solution steps

(w - 3)^{2} = 29 - 5 w

Expand

(w - 3)^{2} : w^{2} - 6 w + 9

w^{2} - 6 w + 9 = 29 - 5 w

Move

5 w

to the left side

w^{2} - w + 9 = 29

Move

29

to the left side

w^{2} - w - 20 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 20) = 9

w_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 1}, w_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 1}

w = \frac{- ( - 1 ) + 9}{2 \cdot 1} : 5

w = \frac{- ( - 1 ) - 9}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

w = 5, w = - 4

- x^{2} - 14 x + 21 = - 2 x + 7

(3 x + 4)^{2} = 0

x, (x + y)^{2} y = 9

x^{2} + 2 = 3 x^{2} - 16

3 x (x - 2) = 2 (x - 2)

What are the solutions to the equation (w-3)^2=29-5w ?
The solutions to the equation (w-3)^2=29-5w are w=5,w=-4
Find the zeros of (w-3)^2=29-5w
The zeros of (w-3)^2=29-5w are w=5,w=-4