What are the solutions to the equation 4y^2+28y=0 ?

The solutions to the equation 4y^2+28y=0 are y=0,y=-7

Find the zeros of 4y^2+28y=0

The zeros of 4y^2+28y=0 are y=0,y=-7

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4 y^{2} + 28 y = 0

y = 0, y = - 7

Solution steps

4 y^{2} + 28 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 0 = 28

y_{1, 2} = \frac{- 28 \pm 28}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 28 + 28}{2 \cdot 4}, y_{2} = \frac{- 28 - 28}{2 \cdot 4}

y = \frac{- 28 + 28}{2 \cdot 4} : 0

y = \frac{- 28 - 28}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

y = 0, y = - 7

\frac{x + 7}{12} - \frac{x ^{2} + 1}{4} = 1 - \frac{x ^{2} + 2}{3}

- t > 0

\frac{8}{- 4}

\frac{5 !}{2 ! ( 5 - 2 ) !}

\frac{25}{49}

What are the solutions to the equation 4y^2+28y=0 ?
The solutions to the equation 4y^2+28y=0 are y=0,y=-7
Find the zeros of 4y^2+28y=0
The zeros of 4y^2+28y=0 are y=0,y=-7