What are the solutions to the equation 10000v^2+11877=21800v ?

The solutions to the equation 10000v^2+11877=21800v are v= 111/100 ,v= 107/100

Find the zeros of 10000v^2+11877=21800v

The zeros of 10000v^2+11877=21800v are v= 111/100 ,v= 107/100

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10000 v^{2} + 11877 = 21800 v

v = \frac{111}{100}, v = \frac{107}{100}

Decimal

v = 1.11, v = 1.07

Solution steps

10000 v^{2} + 11877 = 21800 v

Move

21800 v

to the left side

10000 v^{2} + 11877 - 21800 v = 0

Divide both sides by

10000

\frac{10000 v ^{2}}{10000} + \frac{11877}{10000} - \frac{21800 v}{10000} = \frac{0}{10000}

Write in the standard form

a x^{2} + b x + c = 0

v^{2} - \frac{109 v}{50} + \frac{11877}{10000} = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - \frac{109}{50} ) \pm ( - \frac{109}{50} ) ^{2} - 4 \cdot 1 \cdot \frac{11877}{10000}}{2 \cdot 1}

(- \frac{109}{50})^{2} - 4 \cdot 1 \cdot \frac{11877}{10000} = \frac{1}{25}

v_{1, 2} = \frac{- ( - \frac{109}{50} ) \pm \frac{1}{25}}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - \frac{109}{50} ) + \frac{1}{25}}{2 \cdot 1}, v_{2} = \frac{- ( - \frac{109}{50} ) - \frac{1}{25}}{2 \cdot 1}

v = \frac{- ( - \frac{109}{50} ) + \frac{1}{25}}{2 \cdot 1} : \frac{111}{100}

v = \frac{- ( - \frac{109}{50} ) - \frac{1}{25}}{2 \cdot 1} : \frac{107}{100}

The solutions to the quadratic equation are:

v = \frac{111}{100}, v = \frac{107}{100}

4 x^{2} + 20 x - 24

3 x^{2} + 15 x - 42

∣ 4 - 4 x ∣ < - 13

P = 300 e^{- 0.2 \cdot 17}

15 A C - 9 A D - 30 BC + 18 B D

What are the solutions to the equation 10000v^2+11877=21800v ?
The solutions to the equation 10000v^2+11877=21800v are v= 111/100 ,v= 107/100
Find the zeros of 10000v^2+11877=21800v
The zeros of 10000v^2+11877=21800v are v= 111/100 ,v= 107/100