What are the solutions to the equation y=y(4-y) ?

The solutions to the equation y=y(4-y) are y=0,y=3

Find the zeros of y=y(4-y)

The zeros of y=y(4-y) are y=0,y=3

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y = y (4 - y)

y = 0, y = 3

Solution steps

y = y (4 - y)

Expand

y (4 - y) : 4 y - y^{2}

y = 4 y - y^{2}

Switch sides

4 y - y^{2} = y

Move

y

to the left side

- y^{2} + 3 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

3^{2} - 4 (- 1) \cdot 0 = 3

y_{1, 2} = \frac{- 3 \pm 3}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 3 + 3}{2 ( - 1 )}, y_{2} = \frac{- 3 - 3}{2 ( - 1 )}

y = \frac{- 3 + 3}{2 ( - 1 )} : 0

y = \frac{- 3 - 3}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

y = 0, y = 3

x^{2} - 10 x = - 22

a = (x - 6) (x - 20) - (x - 3) 2 + 50

(π t - 4)^{2} = 4.1 t - 3

- 6 x = - 9 x^{2} + 8

k^{2} + 10 k + 16 = 0

What are the solutions to the equation y=y(4-y) ?
The solutions to the equation y=y(4-y) are y=0,y=3
Find the zeros of y=y(4-y)
The zeros of y=y(4-y) are y=0,y=3