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Popular >

4v^2+28v=0

Solution

4 v^{2} + 28 v = 0

Solution

v = 0, v = - 7

Solution steps

4 v^{2} + 28 v = 0

Solve with the quadratic formula

v_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

v = 0, v = - 7

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Frequently Asked Questions (FAQ)

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