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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 v^{2} + 28 v = 0

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

3 x^{2} + 3 x

\frac{1}{a b ^{2}} + \frac{1}{a ^{2} b}

\frac{8 x ^{- 3} y \cdot 5 x ^{5} y}{10 x ^{6} y ^{2}}

b^{2} - 12 b + 36

p^{2} + 4 p = 1

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