What are the solutions to the equation 4v^2-28v=0 ?

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

Find the zeros of 4v^2-28v=0

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

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4 v^{2} - 28 v = 0

v = 7, v = 0

Solution steps

4 v^{2} - 28 v = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 28)^{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} : 7

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

The solutions to the quadratic equation are:

v = 7, v = 0

\frac{x ^{7} y}{x ^{5} y ^{4}}

2 t^{2} + 7 t - 4 = 0

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(6^{5})^{5}

∣ 3 x - 2 ∣ - ∣ x + 10 ∣ < x

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