What are the solutions to the equation 4v^2+20v=0 ?

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

Find the zeros of 4v^2+20v=0

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

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4 v^{2} + 20 v = 0

v = 0, v = - 5

Solution steps

4 v^{2} + 20 v = 0

Solve with the quadratic formula

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

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

v_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 20 + 20}{2 \cdot 4}, v_{2} = \frac{- 20 - 20}{2 \cdot 4}

v = \frac{- 20 + 20}{2 \cdot 4} : 0

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

The solutions to the quadratic equation are:

v = 0, v = - 5

x^{2} + 16 x + 16 = 0

4 x^{2} + 6 x + 9 = 3 x^{2}

\frac{x ^{2}}{6} = 38

5 x^{2} - 2 x - 16 = 0

(x - 8)^{2} = 7

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