What are the solutions to the equation 5y^2+20y=0 ?

The solutions to the equation 5y^2+20y=0 are y=0,y=-4

Find the zeros of 5y^2+20y=0

The zeros of 5y^2+20y=0 are y=0,y=-4

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5 y^{2} + 20 y = 0

y = 0, y = - 4

Solution steps

5 y^{2} + 20 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

2 0^{2} - 4 \cdot 5 \cdot 0 = 20

y_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 20 + 20}{2 \cdot 5}, y_{2} = \frac{- 20 - 20}{2 \cdot 5}

y = \frac{- 20 + 20}{2 \cdot 5} : 0

y = \frac{- 20 - 20}{2 \cdot 5} : - 4

The solutions to the quadratic equation are:

y = 0, y = - 4

(x + 8) (3 x - 9) = 0

2 = x^{2} - x

a, 2 a^{2} = b^{2}

5 x^{2} - 10 x + 2

\frac{1}{4} x^{2} - 2 x = 0

What are the solutions to the equation 5y^2+20y=0 ?
The solutions to the equation 5y^2+20y=0 are y=0,y=-4
Find the zeros of 5y^2+20y=0
The zeros of 5y^2+20y=0 are y=0,y=-4