What are the solutions to the equation 0=y^2-4y ?

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

Find the zeros of 0=y^2-4y

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

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0 = y^{2} - 4 y

y = 4, y = 0

Solution steps

0 = y^{2} - 4 y

Switch sides

y^{2} - 4 y = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 4}{2 \cdot 1} : 4

y = \frac{- ( - 4 ) - 4}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

y = 4, y = 0

(\frac{3}{2} x - \frac{1}{2})^{2} - 0.8 (1 - x) = 0

(200 + 20 x) (40 - 2 x) = 9000

t = \frac{1}{2} \cdot (- 9.81) \cdot t^{2} + 20 \cdot t + 5

- 4 d^{2} - d + 3 = 0

40 x = - 1000 + 25 x + \frac{1}{2} (1.5) x^{2}

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