What are the solutions to the equation y^2-4y=2y-y^2 ?

The solutions to the equation y^2-4y=2y-y^2 are y=3,y=0

Find the zeros of y^2-4y=2y-y^2

The zeros of y^2-4y=2y-y^2 are y=3,y=0

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y^{2} - 4 y = 2 y - y^{2}

y = 3, y = 0

Solution steps

y^{2} - 4 y = 2 y - y^{2}

Move

y^{2}

to the left side

2 y^{2} - 4 y = 2 y

Move

2 y

to the left side

2 y^{2} - 6 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 6)^{2} - 4 \cdot 2 \cdot 0 = 6

y_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 2}, y_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 2}

y = \frac{- ( - 6 ) + 6}{2 \cdot 2} : 3

y = \frac{- ( - 6 ) - 6}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

y = 3, y = 0

\frac{4 x + 2}{5} \leq 2

∣ 8 y - 4 ∣ > 10

x^{2} + 4 x < 45

7 \cdot \frac{2}{7}

x^{2} \leq 25

What are the solutions to the equation y^2-4y=2y-y^2 ?
The solutions to the equation y^2-4y=2y-y^2 are y=3,y=0
Find the zeros of y^2-4y=2y-y^2
The zeros of y^2-4y=2y-y^2 are y=3,y=0